Question 14 Marks
If $\text{x}^2+\frac{1}{\text{x}^2}=98,$ find the value of $\text{x}^3+\frac{1}{\text{x}^3}.$
AnswerGiven, $\text{x}^2+\frac{1}{\text{x}^2}=98$
We know that$, (x + y)^2 = x^2 + y^2 + 2xy ...(1)$
Substitute $\text{x}^2+\frac{1}{\text{x}^2}=98$ in $eq.(1)$
$\Big(\text{x}+\frac{1}{\text{x}}\Big)^2=\text{x}^2+\frac{1}{\text{x}^2}+2\times\text{x}\times\frac{1}{\text{x}}$
$\Big(\text{x}+\frac{1}{\text{x}}\Big)^2=\text{x}^2+\frac{1}{\text{x}^2}+2$
$\Big(\text{x}+\frac{1}{\text{x}}\Big)^2=98+2$
$\Big(\text{x}+\frac{1}{\text{x}}\Big)^2=100$
$\Big(\text{x}+\frac{1}{\text{x}}\Big)^2=\sqrt{100}$
$\Big(\text{x}+\frac{1}{\text{x}}\Big)^2=\pm10$
We need to find $\text{x}^3+\frac{1}{\text{x}^3}$
So$, a^3 + b^3 = (a - b)(a^2 + b^2 - ab)$
$\text{x}^3+\frac{1}{\text{x}^3}=\Big(\text{x}+\frac{1}{\text{x}}\Big)\Big(\text{x}^2+\frac{1}{\text{x}^2}\Big)-\Big(\text{x}\times\frac{1}{\text{x}}\Big)$
We know that,
$\Big(\text{x}+\frac{1}{\text{x}}\Big)=10$ and $\Big(\text{x}^2+\frac{1}{\text{x}^2}\Big)=98$
$\Rightarrow\text{x}^3+\frac{1}{\text{x}^3}=10(98-1)$
$\Rightarrow\text{x}^3+\frac{1}{\text{x}^3}=10(97)$
$\Rightarrow\text{x}^3+\frac{1}{\text{x}^3}=970$
Hence, the value of $\text{x}^3+\frac{1}{\text{x}^3}=970.$
View full question & answer→Question 24 Marks
If $\text{x}+\frac{1}{\text{x}}=\sqrt{5}$ find the value of $\text{x}^2+\frac{1}{\text{x}^2}$ and $\text{x}^4+\frac{1}{\text{x}^4}$
AnswerWe have,
$\Big(\text{x}+\frac{1}{\text{x}}\Big)^2=\text{x}^2+\frac{1}{\text{x}^2}+2\times\text{x}\times\frac{1}{\text{x}}$
$\Rightarrow\Big(\text{x}+\frac{1}{\text{x}}\Big)^2=\text{x}^2+\frac{1}{\text{x}^2}+2$
$\Rightarrow(\sqrt{5})^2=\text{x}^2+\frac{1}{\text{x}^2}+2$ $\big[\because\text{x}+\frac{1}{\text{x}}=\sqrt{5}\big]$
$\Rightarrow5=\text{x}^2+\frac{1}{\text{x}^2}+2$
$\Rightarrow\ \text{x}^2+\frac{1}{\text{x}^2}=3....(\text{i})$
Now, $\Big(\text{x}^2+\frac{1}{\text{x}^2}\Big)^2=\text{x}^4+\frac{1}{\text{x}^4}+2\times\text{x}^2\times\frac{1}{\text{x}^2}$
$\Rightarrow\Big(\text{x}^2+\frac{1}{\text{x}^2}\Big)^2=\text{x}^4+\frac{1}{\text{x}^4}+2$
$\Rightarrow9=\text{x}^4+\frac{1}{\text{x}^4}+2$ $\big[\because\ \text{x}^2+\frac{1}{\text{x}^2}=3\big]$
$\Rightarrow\text{x}^4+\frac{1}{\text{x}^4}=7$
Hence, $\text{x}^2+\frac{1}{\text{x}^2}=3,\ \text{x}^4+\frac{1}{\text{x}^4}=7.$
View full question & answer→Question 34 Marks
If $\text{x}-\frac{1}{\text{x}}=\frac{1}{2},$ then write the value of $4\text{x}^2+\frac{4}{\text{x}^2}.$
AnswerWe have to find the value of $4\text{x}^2+\frac{4}{\text{x}^2}$
Given $\text{x}-\frac{1}{\text{x}}=\frac{1}{2}$
Using identity $(a - b)^2 = a^2 - 2ab + b^2$
Here $\text{a} = \text{x},\ \text{b}=\frac{1}{\text{x}}$
$\Big(\text{x}-\frac{1}{\text{x}}\Big)^2=\text{x}^2-2\times\text{x}\times\frac{1}{\text{x}}+\Big(\frac{1}{\text{x}}\Big)^2$
$\Big(\text{x}-\frac{1}{\text{x}}\Big)^2=\text{x}^2-2\times\not\text{x}\times\frac{1}{\not\text{x}}+\frac{1}{\text{x}}\times\frac{1}{\text{x}}$
$\Big(\text{x}-\frac{1}{\text{x}}\Big)^2=\text{x}^2-2+\frac{1}{\text{x}^2}$
By substituting the value of $\text{x}-\frac{1}{\text{x}}=\frac{1}{2}$ we get
$\Big(\frac{1}{2}\Big)^2=\text{x}^2+\frac{1}{\text{x}^2}-2$
By transposing -2 to left hand side we get
$\frac{1}{4}+2=\text{x}^2+\frac{1}{\text{x}^2}$
By taking least common multiply we get
$\frac{1}{4}+\frac{2}{1}=\text{x}^2+\frac{1}{\text{x}^2}$
$\frac{1}{4}+\frac{2}{1}\times\frac{4}{4}=\text{x}^2+\frac{1}{\text{x}^2}$
$\frac{1}{4}+\frac{8}{4}=\text{x}^2+\frac{1}{\text{x}^2}$
$\frac{1+8}{4}=\text{x}^2+\frac{1}{\text{x}^2}$
$\frac{+9}{4}=\text{x}^2+\frac{1}{\text{x}^2}$
By multiplying 4 on both sides we get
$4\times\frac{9}{4}=4\text{x}^2+4\times\frac{1}{\text{x}^2}$
$4\times\frac{9}{4}=4\text{x}^2+\frac{4}{\text{x}^2}$
$9 = 4\text{x}^2+\frac{4}{\text{x}^2}$
Hence the value of $4\text{x}^2+\frac{4}{\text{x}^2}$ is $9.$
View full question & answer→Question 44 Marks
Prove that $\text{a}^2+\text{b}^2+\text{c}^2-\text{ab}-\text{bc}-\text{ca}$ is always non negetive for all values of a, b and c.
AnswerIn the given problem, we have to prove $\text{a}^2+\text{b}^2+\text{c}^2-\text{ab}-\text{bc}-\text{ca}$ is always non negetive for all a, b , c that is we have to prove that $\text{a}^2+\text{b}^2+\text{c}^2-\text{ab}-\text{bc}-\text{ca}\geq0$
Consider,
$\text{a}^2+\text{b}^2+\text{c}^2-\text{ab}-\text{bc}-\text{ca}$
$\text{a}^2+\text{b}^2+\text{c}^2-\text{ab}-\text{bc}-\text{ca}\\=\frac{1}{2}\big(2\text{a}^2+2\text{b}^2+2\text{c}^2-2\text{ab}-2\text{bc}-2\text{ca}\big)$
$=\frac{1}{2}[(\text{a}-\text{b}^2)+(\text{b}-\text{c}^2)+(\text{c}-\text{a}^2)]$
View full question & answer→Question 54 Marks
Simplify the following: $\Big(\frac{\text{x}}{2}+\frac{\text{y}}{3}\Big)^3-\Big(\frac{\text{x}}{2}-\frac{\text{y}}{3}\Big)^3$
AnswerGiven $\Big(\frac{\text{x}}{2}+\frac{\text{y}}{3}\Big)^3-\Big(\frac{\text{x}}{2}-\frac{\text{y}}{3}\Big)^3$
We shall use the identity $a^3 - b^3 = (a - b)(a^2 + b^2+ ab)$
Here $\text{a}=\Big(\frac{\text{x}}{2}+\frac{\text{y}}{3}\Big),\ \text{b=}\Big(\frac{\text{x}}{2}-\frac{\text{y}}{3}\Big)$
By applying identity we get
$=\bigg(\Big(\frac{\text{x}}{2}+\frac{\text{y}}{3}\Big)-\Big(\frac{\text{x}}{2}-\frac{\text{y}}{3}\Big)\bigg)$
$\bigg[\Big(\frac{\text{x}}{2}+\frac{\text{y}}{3}\Big)^2+\Big(\frac{\text{x}}{2}-\frac{\text{y}}{3}\Big)^2-\Big(\frac{\text{x}}{2}+\frac{\text{y}}{3}\Big)\Big(\frac{\text{x}}{2}-\frac{\text{y}}{3}\Big)\bigg]$
$=\Big(\frac{\text{x}}{2}+\frac{\text{y}}{3}-\frac{\text{x}}{2}+\frac{\text{y}}{3}\Big)\Bigg[\bigg(\Big(\frac{\text{x}}{2}\Big)^2+\Big(\frac{\text{y}}{3}\Big)^2+\frac{2\text{xy}}{6}\Big)$
$+\bigg(\Big(\frac{\text{x}}{2}\Big)^2+\Big(\frac{\text{y}}{3}\Big)^2-\frac{2\text{xy}}{6}\bigg)+\bigg(\Big(\frac{\text{x}}{2}\Big)^2-\Big(\frac{\text{y}}{3}\Big)^2\bigg)\Bigg]$
$=\frac{2\text{y}}{3}\Big[\Big(\frac{\text{x}^2}{4}+\frac{\text{y}^2}{9}+\frac{2\text{xy}}{6}\Big)+\Big(\frac{\text{x}^2}{4}+\frac{\text{y}^2}{9}-\frac{2\text{xy}}{6}\Big)+\frac{\text{x}^2}{4}-\frac{\text{y}^2}{9}\Big]$
$=\frac{2\text{y}}{3}\Big[\frac{\text{x}^2}{4}+\frac{\text{y}^2}{9}+\frac{2\text{xy}}{6}+\frac{\text{x}^2}{4}+\frac{\text{y}^2}{9}-\frac{2\text{xy}}{6}+\frac{\text{x}^2}{4}-\frac{\text{y}^2}{9}\Big]$
$$By rearranging the variable we get
$=\frac{2\text{y}}{3}\Big[\frac{\text{x}^2}{4}+\frac{\text{y}^2}{9}+\frac{\text{x}^2}{4}+\frac{\text{x}^2}{9}\Big]$
$=\frac{2\text{xy}}{3}\Big[\frac{3\text{x}^2}{4}+\frac{\text{y}^2}{9}\Big]$
$=\frac{\text{x}^2\text{y}}{2}+\frac{2\text{y}^3}{27}$
Hence the simplified value of
$\Big(\frac{\text{x}}{2}+\frac{\text{y}}{3}\Big)^3-\Big(\frac{\text{x}}{2}-\frac{\text{y}}{3}\Big)^3$ is $\frac{\text{x}^2\text{y}}{2}+\frac{2\text{y}^3}{27}.$
View full question & answer→Question 64 Marks
Simplify the following: $(2x - 5y)^3 - (2x + 5y)^3$
AnswerGiven $(2x - 5y)^3 - (2x + 5y)^3$
We shall use the identity $a^3 - b^3 = (a - b)(a^2 + b^2 + ab)$
Here $a = (2x - 5y), b = (2x + 5y)$
By applying the identity we get
$=\big(2\text{x}-5\text{y}-2\text{x}+5\text{y}\big)$
$\Big[(2\text{x}-5\text{y})^2+(2\text{x}+5\text{y})^2\big((2\text{x}-5\text{y})\times(2\text{x}+5\text{y})\big)\Big]$
$=\big(2\text{x}-5\text{y}-2\text{x}-5\text{y})\Big[2\text{x}\times2\text{x}+5\text{y}\times5\text{y}-2\times2\text{x}\times5\text{y}\big)$
$+\big(2\text{x}\times2\text{x}+5\text{y}\times5\text{y}+2\times2\text{x}\times5\text{y}\big)+\big(4\text{x}^2-25\text{y}^2\big)\Big]$
$=(-10\text{y})\Big[\big(4\text{x}^2+25\text{y}^2-20\text{xy}\big)$
$+\big(4\text{x}^2+25\text{y}^2+20\text{xy}\big)+4\text{x}^2-25\text{y}^2\Big]$
$=(-10\text{y})\Big[4\text{x}^2+25\text{y}^2-20\text{xy}+4\text{x}^2+25\text{y}^2+20\text{xy}+4\text{x}^2-25\text{y}^2\Big]$
By rearranging the variable we get,
$=(-10\text{y})\Big[4\text{x}^2+4\text{x}^2+4\text{x}^2+25\text{y}^2\Big]$
$=-10\text{y}\times\big[12\text{x}^2+25\text{y}^2\big]$
$=-120\text{x}^2\text{y}-250\text{y}^3$
Hence the value of $\big(2\text{x}-5\text{y}\big)^3-\big(2\text{x}+5\text{y}\big)^3$ is $-120\text{x}^2\text{y}-250\text{y}^3.$
View full question & answer→Question 74 Marks
If $a - b = 5$ and $ab = 12,$ find the value of $a^2 + b^2.$
AnswerWe have to find the value $a^2 + b^2$
Given $a - b = 5, ab = 12$
Using identity $(a - b)^2 = a^2 - 2ab + b^2$
By substituting the value of $a - b = 5, ab = 12$ we get ,
$(5)^2 = a^2 + b^2 - 2 \times 12$
$5 \times 5 = a^2 + b^2 - 2 \times 12$
$25 = a^2 + b^{2 }- 24$
By transposing $-24$ to left hand side we get
$25 + 24 = a^2 + b^2$
$49 = a^2 + b^2$
Hence the value of a$^2 + b^2$ is $49.$
View full question & answer→Question 84 Marks
If $\text{x}^2+\frac{1}{\text{x}^2}=51,$ find the value of $\text{x}^3-\frac{1}{\text{x}^3}.$
AnswerGiven, $\text{x}^2+\frac{1}{\text{x}^2}=51$
We know that, $(x - y)^2 = x^2 + y^2 - 2xy ...(1)$
Substitute $\text{x}^2+\frac{1}{\text{x}^2}=51$in $eq.(1)$
$\Big(\text{x}-\frac{1}{\text{x}}\Big)^2=\text{x}^2+\frac{1}{\text{x}^2}-2\times\text{x}\times\frac{1}{\text{x}}$
$\Big(\text{x}-\frac{1}{\text{x}}\Big)^2=\text{x}^2+\frac{1}{\text{x}^2}-2$
$\Big(\text{x}-\frac{1}{\text{x}}\Big)^2=51-2$
$\Big(\text{x}-\frac{1}{\text{x}}\Big)^2=49$
$\Big(\text{x}-\frac{1}{\text{x}}\Big)^2=\sqrt{49}$
$\Big(\text{x}-\frac{1}{\text{x}}\Big)^2=\pm7$
We need to find $\text{x}^3-\frac{1}{\text{x}^3}$
So$, a^3 - b^3 = (a - b)(a^2 + b^2 + ab)$
$\text{x}^3-\frac{1}{\text{x}^3}=\Big(\text{x}-\frac{1}{\text{x}}\Big)\Big(\text{x}^2+\frac{1}{\text{x}^2}\Big)+\Big(\text{x}\times\frac{1}{\text{x}}\Big)$
We know that,
$\Big(\text{x}-\frac{1}{\text{x}}\Big)=7$ and $\Big(\text{x}^2+\frac{1}{\text{x}^2}\Big)=51$
$\Rightarrow\text{x}^3-\frac{1}{\text{x}^3}=7(51+1)$
$\Rightarrow\text{x}^3-\frac{1}{\text{x}^3}=7(52)$
$\Rightarrow\text{x}^3-\frac{1}{\text{x}^3}=364$
Hence, the value of $\text{x}^3-\frac{1}{\text{x}^3}=364.$
View full question & answer→Question 94 Marks
Simplify the following products:
$\Big(\text{m}+\frac{\text{n}}{7}\Big)^3\Big(\text{m}-\frac{\text{n}}{7}\Big)$
AnswerIn the given problem, we have to find product of $\Big(\text{m}+\frac{\text{n}}{7}\Big)^3\Big(\text{m}-\frac{\text{n}}{7}\Big)$
We have been given $\Big(\text{m}+\frac{\text{n}}{7}\Big)^3\Big(\text{m}-\frac{\text{n}}{7}\Big)$
On rearranging we get,
$\Big(\text{m}+\frac{\text{n}}{7}\Big)^3\Big(\text{m}-\frac{\text{n}}{7}\Big)=\Big(\text{m}+\frac{\text{n}}{7}\Big)^2\Big(\text{m}+\frac{\text{n}}{7}\Big)\Big(\text{m}-\frac{\text{n}}{7}\Big)$
We shall use the identity $\big(\text{x}-\text{y}\big)\big(\text{x}+\text{y}\big)=\text{x}^2-\text{y}^2$
By substituting $\text{x}=\text{m},\ \text{y}=\frac{\text{n}}{7},$ we get,
$\Big(\text{m}+\frac{\text{n}}{7}\Big)^3\Big(\text{m}-\frac{\text{n}}{7}\Big)$
$=\Big(\text{m}+\frac{\text{n}}{7}\Big)^2\bigg(\text{m}^2-\Big(\frac{\text{n}}{7}\Big)^2\bigg)$
$=\Big(\text{m}+\frac{\text{n}}{7}\Big)^2\Big(\text{m}^2-\frac{\text{n}^2}{49}\Big)$
Hence the value of $\Big(\text{m}+\frac{\text{n}}{7}\Big)^3\Big(\text{m}-\frac{\text{n}}{7}\Big)$ is $\Big(\text{m}+\frac{\text{n}}{7}\Big)^2\Big(\text{m}^2-\frac{\text{n}^2}{49}\Big)$
View full question & answer→Question 104 Marks
If $\text{x}^4+\frac{1}{\text{x}^4}=194,$ find $\text{x}^3+\frac{1}{\text{x}^3},\ \text{x}^2+\frac{1}{\text{x}^2}$ and $\text{x}+\frac{1}{\text{x}}.$
AnswerIn the given problem, we have to find the value of $\text{x}^3+\frac{1}{\text{x}^3},\ \text{x}^2+\frac{1}{\text{x}^2},\ \text{x}+\frac{1}{\text{x}}$
Given $\text{x}^4+\frac{1}{\text{x}^4}=194$
By adding and subtracting $2\times\text{x}^2\times\frac{1}{\text{x}^2}$ in left hand side of $\text{x}^4+\frac{1}{\text{x}^4}=194$ we get,
$\text{x}^4+\frac{1}{\text{x}^4}+2\times\text{x}^2\times\frac{1}{\text{x}^2}-2\times\text{x}^2\times\frac{1}{\text{x}^2}=194$
$\text{x}^4+\frac{1}{\text{x}^4}+2\times\text{x}^2\times\frac{1}{\text{x}^2}-2\times\Big(\not\text{x}^2\times\frac{1}{\not\text{x}^2}\Big)=194$
$\Big(\text{x}^2\times\frac{1}{\text{x}^2}\Big)^2-2=194$
$\Big(\text{x}^2\times\frac{1}{\text{x}^2}\Big)^2=194+2$
$\Big(\text{x}^2\times\frac{1}{\text{x}^2}\Big)^2=196$
$\Big(\text{x}^2\times\frac{1}{\text{x}^2}\Big)^2=(14)^2$
$\Big(\text{x}^2\times\frac{1}{\text{x}^2}\Big)=14$
Again by adding and subtracting $2\times\text{x}\times\frac{1}{\text{x}}$ in left hand side of $\Big(\text{x}^2+\frac{1}{\text{x}^2}\Big)=14$ we get,
$\text{x}^2+\frac{1}{\text{x}^2}+2\times\text{x}\times\frac{1}{\text{x}}-2\times\text{x}\times\frac{1}{\text{x}}=14$
$\Big(\text{x}+\frac{1}{\text{x}}\Big)^2-2\times\not\text{x}\times\frac{1}{\not\text{x}}=14$
$\Big(\text{x}+\frac{1}{\text{x}}\Big)^2-2=14$
$\Big(\text{x}+\frac{1}{\text{x}}\Big)^2=14+2$
$\Big(\text{x}+\frac{1}{\text{x}}\Big)^2=16$
$\Big(\text{x}+\frac{1}{\text{x}}\Big)^2=4\times4$
$\Big(\text{x}+\frac{1}{\text{x}}\Big)=4$
Now cubing on both sides of $\Big(\text{x}+\frac{1}{\text{x}}\Big)=4$ we get
$\Big(\text{x}+\frac{1}{\text{x}}\Big)^3=4^3$
we shall use identity $(a + b)^3 = a^3 + b^3 + 3ab(a + b)$
$\text{x}^3+\frac{1}{\text{x}^3}+3\times\text{x}\times\frac{1}{\text{x}}\Big(\text{x}\times\frac{1}{\text{x}}\Big)=4\times4\times4$
$\text{x}^3+\frac{1}{\text{x}^3}+3\times\not\text{x}\times\frac{1}{\not\text{x}}\times4=64$
$\text{x}^3+\frac{1}{\text{x}^3}+12=64$
$\text{x}^3+\frac{1}{\text{x}^3}=64-12$
$\text{x}^3+\frac{1}{\text{x}^3}=52$
Hence the value of $\text{x}^3+\frac{1}{\text{x}^3},\ \text{x}^2+\frac{1}{\text{x}^2} ,\ \text{x}+\frac{1}{\text{x}}$ is $52, 14, 4$ respectively.
View full question & answer→Question 114 Marks
Find the cube of the following binomial expressions : $\frac{1}{\text{x}}+\frac{\text{y}}{3}$
AnswerGiven,
$\frac{1}{\text{x}}+\frac{\text{y}}{3}$
The above equation is in the form of $(a + b)^{3 }= a^3 + b^3 + 3ab(a + b)$
We know that,$ \text{a} =\frac{1}{\text{x}},\text{b} =\frac{\text{y}}{3}$
By using $(a + b)^{3 }$formula
$\Big(\frac{1}{\text{x}}+\frac{\text{y}}{3}\Big)^3$
$=\Big(\frac{1}{\text{x}}\Big)^3+\Big(\frac{\text{y}}{3}\Big)^3+3\Big(\frac{1}{\text{x}}\Big)\Big(\frac{\text{y}}{3}\Big)\Big(\frac{1}{\text{x}}+\frac{\text{y}}{3}\Big)$
$=\frac{1}{\text{x}^3}+\frac{\text{y}^3}{27}+3\times\frac{1}{\text{x}}\times\frac{\text{y}}{3}\Big(\frac{1}{\text{x}}+\frac{\text{y}}{3}\Big)$
$=\frac{1}{\text{x}^3}+\frac{\text{y}^3}{27}+\frac{\text{y}}{\text{x}}\Big(\frac{1}{\text{x}}+\frac{\text{y}}{3}\Big)$
$=\frac{1}{\text{x}^3}+\frac{\text{y}^3}{27}+\Big(\frac{\text{y}}{\text{x}}\times\frac{1}{\text{x}}\Big)+\Big(\frac{\text{y}}{\text{x}}\times\text{y}^3\Big)$
$=\Big(\frac{1}{\text{x}^3}+\frac{\text{y}^3}{27}+\frac{\text{y}}{\text{x}^2}+\frac{\text{y}^2}{3\text{x}}\Big)$
Hence
$\Big(\frac{1}{\text{x}}+\frac{\text{y}}{3}\Big)^3=\Big(\frac{1}{\text{x}^3}+\frac{\text{y}^3}{27}+\frac{\text{y}}{\text{x}^2}+\frac{\text{y}^2}{3\text{x}}\Big)$
View full question & answer→Question 124 Marks
If $\text{x}+\frac{1}{\text{x}}=3,$ then find the value of $\text{x}^2+\frac{1}{\text{x}^2}.$
AnswerWe have to find the value of $\text{x}^2+\frac{1}{\text{x}^2}$
Given $\text{x}+\frac{1}{\text{x}}=3$
Using identity $(a + b)^2 = a^2 + 2ab + b^2$
Here $\text{a}=\text{x},\ \text{b}=\frac{1}{\text{x}}$
$\Big(\text{x}+\frac{1}{\text{x}}\Big)^2=\text{x}^2+2\times\text{x}\times\frac{1}{\text{x}}+\Big(\frac{1}{\text{x}}\Big)^2$
$\Big(\text{x}+\frac{1}{\text{x}}\Big)^2=\text{x}\times\text{x}+2\times\not{\text{x}}\times\frac{1}{\not{\text{x}}}+\frac{1}{\text{x}}\times\frac{1}{\text{x}}$
$\Big(\text{x}+\frac{1}{\text{x}}\Big)^2=\text{x}^2+2+\frac{1}{\text{x}^2}$
By substituting the value of $\text{x}+\frac{1}{\text{x}}=3$ we get,
$(3)^2=\text{x}^2+2+\frac{1}{\text{x}^2}$
$3\times3=\text{x}^2+2+\frac{1}{\text{x}^2}$
By transposing $+ 2$ to left hand side, we get
$9-2=\text{x}^2+\frac{1}{\text{x}^2}$
$7=\text{x}^2+\frac{1}{\text{x}^2}$
Hence the value of $\text{x}^2+\frac{1}{\text{x}^2}$ is $7.$
View full question & answer→Question 134 Marks
If $a + b = 10$ and $ab = 16$, find the value of $a^2 - ab + b^2$ and $a^2 + ab + b^2.$
AnswerWe have,
$a^2 - ab + b^2 = a^2+ b^2 - ab$
$= a^2 + b^2 - ab + 2ab - 2ab [$Adding and substracting $2ab]$
$= (a^2 + b^2 + 2ab) - 3ab \big[\because (a + b)2 = a2 + b2 - 2ab\big]$
$= (a + b)^2 - 3 \times 16 \big[\because a + b = 10$ and $b = 16 \big]$
$= (10)^2 - 3 \times 16$
$= 100 - 48$
$= 52$
$\Rightarrow a^2 - ab + b^2 = 52$
We have,
$a^2 + ab + b^2 = a^2 + ab + b^2 + ab - ab [$Adding and substracting $ab]$
$= (a^2 + b^2 + 2ab) - ab$
$= (a + b)^2 - ab\big[\because (a + b)2 = a2 + b2 - 2ab\big]$
$= (10)^2 - 16$
$= 100 - 16$
$\Rightarrow a^2 + ab + b^2 = 84$
Hence$, a^2 - ab + b^2 = 52,$ and $a^2 + ab + b^2 = 84.$
View full question & answer→Question 144 Marks
Find the value of $64x^3 - 125z^3,$ if $4x - 5z = 16$ and $xz = 12.$
AnswerGiven, $64x^3 - 125z^3$
Here$, 4x - 5z = 16$ and $xz = 12$
Cubing $4x - 5z = 16$ on both sides
$(4x - 5z)^3 = 16^3$
We know that$, (a - b)^3 = a^3 - b^3 - 3ab(a - b)$
$(4x)^3 - (5z)^3 - 3(4x)(5z)(4x - 5z) = 16^3$
$64x^3 - 125z^3 - 60(xz)(16) = 4096$
$64x^3 - 125z^3 - 60(12)(16) = 4096$
$64x^3 - 125z^3 - 11520 = 4096$
$64x^3 - 125z^3 = 4096 + 11520$
$64x^3 - 125z^3 = 15616$
The value of $64x^3 - 125z^3 $
$= 15616$
View full question & answer→Question 154 Marks
If $\text{a}^2-\frac{1}{\text{a}^2}=102,$ find the value of $\text{a}-\frac{1}{\text{a}}.$
AnswerWe have to find the value of $\text{a}-\frac{1}{\text{a}}$
Given $\text{a}^2-\frac{1}{\text{a}^2}=102$
Using identity $(x - y)^2 = x^2 + y^2 - 2xy$
Here $\text{x}=\text{a},\ \text{y}=\frac{1}{\text{a}}$
$\Big(\text{a}-\frac{1}{\text{a}}\Big)^2=\text{a}^2+\Big(\frac{1}{\text{a}}\Big)^2-2\times\text{a}\times\frac{1}{\text{a}}$
$\Big(\text{a}-\frac{1}{\text{a}}\Big)^2=\text{a}^2+\frac{1}{\text{a}^2}-2\times\not\text{a}\times\frac{1}{\not\text{a}}$
By substituting $\text{a}^2-\frac{1}{\text{a}^2}=102$ we get
$\Big(\text{a}-\frac{1}{\text{a}}\Big)^2=102-2$
$\Big(\text{a}-\frac{1}{\text{a}}\Big)^2=100$
$\Big(\text{a}-\frac{1}{\text{a}}\Big)\Big(\text{a}-\frac{1}{\text{a}}\Big)=10\times10$
$\Big(\text{a}-\frac{1}{\text{a}}\Big)=10$
Hence the value of $\Big(\text{a}-\frac{1}{\text{a}}\Big)$ is $10$.
View full question & answer→Question 164 Marks
Find the value of $27x^3 + 8y^3$, if $3x + 2y = 14$ and $xy = 8$
AnswerIn the given problem, we have to find the value of $27x^3 + 8y^3$
Given $3x + 2y = 14, xy = 8$
On cubing both sides we get,
$(3x + 2y)^3 = (14)^3$
We shall use identity $(a + b)^3 = a^3 + b^3+ 3ab(a + b)$
$27x^3 + 8y^3 + 3(3x)(2y)(3x + 2y) = 14 \times 14 \times 14$
$27x^3 + 8y^3 + 18(xy)(3x + 2y) = 14 \times 14 \times 14$
$27x^3 + 8y^3 + 18(8)(14) = 2744$
$27x^3 + 8y^3 + 2016 = 2744$
$27x^3 + 8y^3 = 2744 - 2016$
$27x^3 + 8y^3= 728$
Hence the value of $27x^3 + 8y^3$ is $728.$
View full question & answer→Question 174 Marks
If $2x + 3y = 13$ and $xy = 6,$ find the value of $8x^3 + 27y^3.$
AnswerGiven, $2x + 3y = 13, xy = 6$
We know that,
$(2x + 3y)^3 = 13^2$
$\Rightarrow 8x^3 + 27y^3 + 3(2x)(3y)(2x + 3y) = 2197$
$\Rightarrow 8x^3 + 27y^3 + 18xy(2x + 3y) = 2197$
Substitute $2x + 3y = 13, xy = 6$
$\Rightarrow 8x^3 + 27y^3 + 18(6)(13) = 2197$
$\Rightarrow 8x^3 + 27y^3 + 1404 = 2197$
$\Rightarrow 8x^3 + 27y^3 = 2197 – 1404$
$\Rightarrow 8x^3 + 27y^3 = 793$
Hence, the value of $8x^3 + 27y^3 $
$= 793$
View full question & answer→Question 184 Marks
If $\text{x}+\frac{1}{\text{x}}=5,$ find the value of $\text{x}^3+\frac{1}{\text{x}^3}.$
AnswerGiven, $\text{x}+\frac{1}{\text{x}}=5$
We know that, $(a + b)^3 = a^3 + b^3 + 3ab(a + b) ...(1)$
Substitute $\text{x}+\frac{1}{\text{x}}=5$ in $eq(1)$
$$$\Rightarrow\Big(\text{x}+\frac{1}{\text{x}}\Big)^3=\text{x}^3+\frac{1}{\text{x}^3}+3\Big(\text{x}\times\frac{1}{\text{x}}\Big)\Big(\text{x}+\frac{1}{\text{x}}\Big)$
$\Rightarrow5^3=\text{x}^3+\frac{1}{\text{x}^3}+3\Big(\text{x}\times\frac{1}{\text{x}}\Big)\Big(\text{x}+\frac{1}{\text{x}}\Big)$
$\Rightarrow125=\text{x}^3+\frac{1}{\text{x}^3}+3\Big(\text{x}+\frac{1}{\text{x}}\Big)$
$\Rightarrow125=\text{x}^3+\frac{1}{\text{x}^3}+3(5)$
$\Rightarrow125=\text{x}^3+\frac{1}{\text{x}^3}+15$
$\Rightarrow125-15=\text{x}^3+\frac{1}{\text{x}^3}$
$\Rightarrow\text{x}^3+\frac{1}{\text{x}^3}=110$
Hence, the result is $\text{x}^3+\frac{1}{\text{x}^3}=110.$
View full question & answer→Question 194 Marks
If $3x - 7y = 10$ and $xy = -1$, find the value of $9x^2 + 49y^2.$
AnswerIn the given problem, we have to find $9x^2 + 49y^2$
We have been given $3x - 7y = 10$ and $xy = -1$
Let us take $3x - 7y = 10$
On squaring both side we get,
$(3x - 7y)^2 = (10)^2$
$(3x \times 3x + 7y \times 7y - 2 \times 3x \times 7y) = 100$
We shall use the identity $(x - y)^2$
$= x^2 + 2xy + y^2$
$9x^2 + 49y^2 - 42xy = 100$
$9x^2 + 49y^2 - 42(xy) = 100$
By substituting $xy = -1$ we get
$9x^2 + 49y^2 - 42(-1) = 100$
$9x^2 + 49y^2 + 42 = 100$
$9x^2 + 49y^2 = 100 - 42$
$9x^2 + 49y^2 = 58$
Hence the value of $9x^2 + 49y^2$ is $58.$
View full question & answer→Question 204 Marks
If $\text{x}-\frac{1}{\text{x}}=3+2\sqrt2,$ find the value of $\text{x}^3-\frac{1}{\text{x}^3}.$
AnswerIn the given problem, we have to find the value of $\text{x}^3-\frac{1}{\text{x}^3}$
Given $\text{x}-\frac{1}{\text{x}}=3+2\sqrt2,$
Cubing on both side $\text{x}-\frac{1}{\text{x}}=3+2\sqrt2,$ we get
$\Big(\text{x}-\frac{1}{\text{x}}\Big)^3=\Big(3+2\sqrt2\Big)^3$
We shall use identity $(a + b)^3 = a^3 + b^3 + 3ab(a + b)$
$\big(3+2\sqrt2\big)^3$
$=\text{x}^3-\frac{1}{\text{x}^3}-3\times\text{x}\times\frac{1}{\text{x}}\Big(\text{x}-\frac{1}{\text{x}}\Big)$
$3^3+\big(2\sqrt2\big)+3\times3\times2\sqrt2\big(3+2\sqrt2\big)$
$=\text{x}^3-\frac{1}{\text{x}^3}-3\times\not\text{x}\times\frac{1}{\not\text{x}}\times\big(3+2\sqrt2\big)$
$27+16\sqrt2+18\sqrt2\big(3+2\sqrt2\big)$
$=\text{x}^3-\frac{1}{\text{x}^3}-3\big(3+2\sqrt2\big)$
$27+16\sqrt2+18\sqrt2\times3+18\sqrt2\times2\sqrt2$
$=\text{x}^3-\frac{1}{\text{x}^3}-9-6\sqrt2$
$27+16\sqrt2+54\sqrt2+72=\text{x}^3-\frac{1}{\text{x}^3}-9-6\sqrt2$
$27+16\sqrt2+54\sqrt2+72+9+6\sqrt2=\text{x}^3-\frac{1}{\text{x}^3}$
$\big[27+72+9\big]+\big[16\sqrt2+54\sqrt2+6\sqrt2\big]=\text{x}^3-\frac{1}{\text{x}^3}$
$108+76\sqrt2=\text{x}^3-\frac{1}{\text{x}^3}$
Hence the value of $\text{x}^3-\frac{1}{\text{x}^3}$ is $108+76\sqrt2.$
View full question & answer→Question 214 Marks
Find the cube of the following binomial expressions: $2\text{x}+\frac{3}{\text{x}}$
AnswerGiven,
$2\text{x}+\frac{3}{\text{x}}$
The above equation is in the form of $(a + b)^{3 }= a^3 + b^3 + 3ab(a + b)$
we know that $\text{a}=2\text{x},\text{b}=\frac{3}{\text{x}}$
By using $(a + b)^{3 }$formula
$=8\text{x}^3+\frac{27}{\text{x}^3}+\frac{18\text{x}}{\text{x}}\Big(\frac{2}{\text{x}}+\frac{3}{\text{x}}\Big)$
$=8\text{x}^3+\frac{27}{\text{x}^3}+\frac{18\text{x}}{\text{x}}\Big(2{\text{x}}+\frac{3}{\text{x}}\Big)$
$=8\text{x}^3+\frac{27}{\text{x}^3}+\Big({18}\times2{\text{x}}\Big)+\Big(18\times\frac{3}{\text{x}}\Big)$
$=\Big(8\text{x}^3+\frac{27}{\text{x}^3}+36\times\frac{54}{\text{x}}\Big)$
Hence
The cube of $\Big(2\text{x}+\frac{3}{\text{x}}\Big)^3=\Big(8\text{x}^3+\frac{27}{\text{x}^3}+36\times\frac{54}{\text{x}}\Big)$
View full question & answer→Question 224 Marks
Simplify: $(x^2 + y^2 - z^2)^2 - (x^2 - y^2 + z^2)^2$
AnswerWe have $(x^2 + y^2 - z^2)^2 - (x^2 - y^2 + z^2)^2$
Using formula $(x + y + z)^2$
$= x^2 + y^2 + z^2 + 2xy + 2yz + 2zx,$ we get
$(x^2 + y^2 - z^2)^2 - (x^2 - y^2 + z^2)^2$
$=(\text{x}^2)^2+(\text{y}^2)^2+(-\text{z}^2)^2+2(\text{x}^2)(\text{y}^2)+2(\text{y}^2)(-\text{z}^2)+2(-\text{z}^2)(\text{x}^2)$
$-\Big[(\text{x}^2)^2+(-\text{y}^2)^2+(\text{z}^2)^2+2(\text{x}^2)(-\text{y}^2)+2(-\text{y}^2)(\text{z}^2)+2(\text{z}^2)(\text{x}^2)\Big]$
$=\text{x}^4+\text{y}^4+\text{z}^4+2\text{x}^2\text{y}^2-2\text{y}^2\text{z}^2-2\text{z}^2\text{x}^2$
$-\big[\text{x}^4+\text{y}^4+\text{z}^4-2\text{x}^2\text{y}^2-2\text{y}^2\text{z}^2+2\text{z}^2\text{x}^2\big]$
By canceling the opposite terms, we get
$\big(\text{x}^2 +\text{y}^2 -\text{z}^2\big)^2 -\big(\text{x}^2 -\text{y}^2+\text{z}^2)^2=\not\text{x}^4+\not\text{y}^4+\not\text{z}^4$
$+2\text{x}^2\text{y}^2-2\text{y}^2\text{z}^2-2\text{z}^2\text{x}^2\not\text{x}^4-\not\text{y}^4-\not\text{z}^4+2\text{x}^2\text{y}^2+2\text{y}^2\text{z}^2-2\text{z}^2\text{x}^2$
$=4\text{x}^2\text{y}^2-4\text{z}^2\text{x}^2$
Talking 4x^2 as common factor we get
$\big(\text{x}^2 +\text{y}^2 -\text{z}^2\big)^2 -\big(\text{x}^2 -\text{y}^2+\text{z}^2)^2=4\text{x}^2(\text{y}^2-\text{z}^2)$
Hence the value of $\big(\text{x}^2 +\text{y}^2 -\text{z}^2\big)^2 -\big(\text{x}^2 -\text{y}^2+\text{z}^2)^2$ is $4\text{x}^2(\text{y}^2-\text{z}^2).$
View full question & answer→Question 234 Marks
Simplify: $(a + b + c)^{2 }- (a - b + c)^2$
AnswerIn the given problem, we have to simplify the expressions
Given $(a + b + c)^2 - (a - b + c)^2$
By using identity $(x + y + z)^2 = x^2 + y^2 + z^2 + 2xy + 2yz + 2zx$
Hence the equation becomes
$\big(\text{a}+\text{b}+\text{c}\big)-\big(\text{a}-\text{b}+\text{c}\big)$
$=\Big[\text{a}^2+\text{b}^2+\text{c}^2+2\text{ab}+2\text{bc}+2\text{ca}\Big]$
$-\Big[\text{a}^2+(-\text{b})^2+\text{c}^2+2\text{a}(-\text{b})+2(-\text{b})(\text{c})+2\text{ca}\Big]$
$=\text{a}^2+\text{b}^2+\text{c}^2+2\text{ab}+2\text{bc}+2\text{ca}$
$-\text{a}^2-\text{b}^2-\text{c}^2+2\text{ab}+2\text{bc}-2\text{ca}$
$=\not\text{a}^2-\not\text{a}^2+\not\text{b}^2-\not\text{b}^2+\not\text{c}^2-\not\text{c}^2$
$+2\text{ab}+2\text{ab}+2\text{bc}+2\text{bc}+2\text{ca}-2\text{ca}$
$=4\text{ab}+4\text{bc}$
Talking 4 as common factor we get
$=4\big(\text{ab}+\text{bc}\big)$
Hence the simplified value of $\big(\text{a}+\text{b}+\text{c}\big)^2-\big(\text{a}-\text{b}+\text{c}\big)^2$ is $4\big(\text{ab}+\text{bc}\big).$
View full question & answer→Question 244 Marks
Simplify: $(2x + p - c)^2 - (2x - p + c)^2$
AnswerGiven $(2x + p - c)^2 - (2x - p + c)^2$
By using identity $(x + y + z) = x^2 + y^2 + z^2 + 2xy + 2yz + 2zx,$ we get
$(2x + p - c)^2 - (2x - p + c)^2$
$=(2\text{x})^2+(\text{p})^2+(-\text{c})^2+2(2\text{x})(\text{p})+2(\text{p})(-\text{c})+2(-\text{c})(2\text{x})$
$-\big[(2\text{x})^2+(-\text{p})^2+(\text{c})^2+2(2\text{x})(-\text{p})+2(-\text{p})(\text{c})+2(\text{c})(2\text{x})\big]$
$$$=4\text{x}^2+\text{p}^2+\text{c}^2+4\text{xp}-2\text{cp}-4\text{cx}$
$-\big[4\text{x}^2+\text{p}^2+\text{c}^2-4\text{xp}-2\text{cp}+4\text{cx}\big]$
By cancelling the opposite terms, we get
$\big(2\text{x}+\text{p}-\text{c}\big)^2-\big(2\text{x}-\text{p}+\text{c}\big)^2$
$=4\text{x}^2+\not\text{p}^2+\not\text{c}^2+4\text{xp}-2\text{cp}-4\text{cx}-4\text{x}^2$
$-\not\text{p}^2-\not\text{c}^2+4\text{xp}+2\text{cp}-4\text{cx}$
$=4\text{xp}+4\text{xp}-4\text{cx}-4\text{cx}$
$=8\text{xp}-8\text{cx}$
Talking 8x as common a factor we get,
$\big(2\text{x}+\text{p}-\text{c}\big)^2-\big(2\text{x}-\text{p}+\text{c}\big)^2=8\text{x}(\text{p}-\text{c})$
Hence the value of $\big(2\text{x}+\text{p}-\text{c}\big)^2-\big(2\text{x}-\text{p}+\text{c}\big)^2$ is $8\text{x}(\text{p}-\text{c}).$
View full question & answer→Question 254 Marks
If $\text{x}^2+\frac{1}{\text{x}^2}=79,$ find the value of $\text{x}+\frac{1}{\text{x}}.$
AnswerIn the given problem, we have to find $\Big(\text{x}+\frac{1}{\text{x}}\Big)$
Given $\text{x}^2+\frac{1}{\text{x}^2}=79$
Adding and subtracting 2 on left hand side,
$\text{x}^2+\frac{1}{\text{x}^2}+2-2=79$
$\Big(\text{x}^2+\frac{1}{\text{x}^2}+2\times\text{x}\times\frac{1}{\text{x}}\Big)-2=79$
$\Big(\text{x}+\frac{1}{\text{x}}\Big)^2-2=79$
$\Big(\text{x}+\frac{1}{\text{x}}\Big)^2=79+2$
$\Big(\text{x}+\frac{1}{\text{x}}\Big)^2=81$
$\Big(\text{x}+\frac{1}{\text{x}}\Big)^2=\sqrt{81}$
$\text{x}+\frac{1}{\text{x}}=\sqrt{9\times9}$
$\text{x}+\frac{1}{\text{x}}=\pm9$
Hence the value of $\Big(\text{x}+\frac{1}{\text{x}}\Big)$ is $\pm9$
View full question & answer→Question 264 Marks
Simplify the following products:
$\Big(\frac{1}{2}\text{a}-3\text{b}\Big)\Big(3\text{b}+\frac{1}{2}\text{a}\Big)\Big(\frac{1}{4}\text{a}^2+9\text{b}^2\Big)$
AnswerIn the given problem, we have to find product of $\Big(\frac{1}{2}\text{a}-3\text{b}\Big)\Big(3\text{b}+\frac{1}{2}\text{a}\Big)\Big(\frac{1}{4}\text{a}^2+9\text{b}^2\Big)$
We have been given $\Big(\frac{1}{2}\text{a}-3\text{b}\Big)\Big(3\text{b}+\frac{1}{2}\text{a}\Big)\Big(\frac{1}{4}\text{a}^2+9\text{b}^2\Big)$
On rearranging we get, $\Big(\frac{1}{2}\text{a}-3\text{b}\Big)\Big(\frac{1}{2}\text{a}+3\text{b}\Big)\Big(\frac{1}{4}\text{a}^2+9\text{b}^2\Big)$
We shall use the identity $\big(\text{x}-\text{y}\big)\big(\text{x}+\text{y}\big)=\text{x}^2-\text{y}^2$
By substituting $\text{x}=\frac{1}{2}\text{a},\ \text{y}=3\text{b}$ we get,
$$$\Big(\frac{1}{2}\text{a}-3\text{b}\Big)\Big(3\text{b}+\frac{1}{2}\text{a}\Big)\Big(\frac{1}{4}\text{a}^2+9\text{b}^2\Big)\\=\Big(\frac{1}{2}\text{a}\Big)^2-(3\text{b})^2\Big(\frac{1}{4}\text{a}^2+9\text{b}^2\Big)$
$=\Big(\frac{1}{4}\text{a}^2-9\text{b}^2\Big)\Big(\frac{1}{4}\text{a}^2+9\text{b}^2\Big)$
We shall use the identity $\big(\text{x}-\text{y}\big)\big(\text{x}+\text{y}\big)=\text{x}^2-\text{y}^2$
$\Big(\frac{1}{2}\text{a}-3\text{b}\Big)\Big(3\text{b}+\frac{1}{2}\text{a}\Big)\Big(\frac{1}{4}\text{a}^2+9\text{b}^2\Big)\\=\Big(\frac{1}{4}\text{a}^2\Big)^2-(9\text{b}^2)^2$
$=\frac{1}{16}\text{a}^4-81\text{b}^4$
Hence the value of $\Big(\frac{1}{2}\text{a}-3\text{b}\Big)\Big(3\text{b}+\frac{1}{2}\text{a}\Big)\Big(\frac{1}{4}\text{a}^2+9\text{b}^2\Big)$ is $\frac{1}{16}\text{a}^4-81\text{b}^4.$
View full question & answer→Question 274 Marks
Simplify the following: $(x + 3)^3 + (x - 3)^3$
AnswerIn the given problem, we have to simplify equation
Given $(x + 3)^3 + (x - 3)^3$
We shall use the identity $a^3 + b^3 = (a + b)(a^2 + b^2 - ab)$
Here $a= (x + 3), b = (x - 3)$
By applying identity we get
$=\big(\text{x}+\not3+\text{x}-\not3\big)\big[(\text{x}+3)^2+(\text{x}-3)^2-(\text{x}+3)(\text{x}-3)\big]$
$=2\text{x}\big[\big(\text{x}^2+3^2+2\times\text{x}\times3\big)+\big(\text{x}^2+3^2-2\times\text{x}\times3\big)-(\text{x}^2)-3^2\big]$
$=2\text{x}\big[\big(\text{x}^2+9+6\text{x}\big)+\big(\text{x}^2+9-6\text{x}\big)-\big(\text{x}^2-3^2\big)\big]$
$=2\text{x}\big[\text{x}^2+9+6\text{x}+\text{x}^2+9-6\text{x}-\text{x}^2+9\big]$
$=2\text{x}\big[\text{x}^2+\not\text{x}^2-\not\text{x}^2-6\text{x}+6\text{x}+9+9+9\big]$
$=2\text{x}\big[\text{x}^2+27\big]$
$=2\text{x}^2+54\text{x}$
Hence simplified form of expression $(\text{x}+3)^3+(\text{x}-3)^3$ is $2\text{x}^2+54\text{x}.$
View full question & answer→Question 284 Marks
If $\text{x}-\frac{1}{\text{x}}=5,$ find the value of $\text{x}^3-\frac{1}{\text{x}^3}.$
AnswerGiven,If $\text{x}-\frac{1}{\text{x}}=5$
We know that, $(a - b)^3 = a^3 - b^3 - 3ab(a - b) ...(1)$
Substitute $\text{x}-\frac{1}{\text{x}}=5$ in $eq(1)$
$\Rightarrow\Big(\text{x}-\frac{1}{\text{x}}\Big)^3=\text{x}^3-\frac{1}{\text{x}^3}-3\Big(\text{x}\times\frac{1}{\text{x}}\Big)\Big(\text{x}-\frac{1}{\text{x}}\Big)$
$\Rightarrow5^3=\text{x}^3-\frac{1}{\text{x}^3}-3\Big(\text{x}-\frac{1}{\text{x}}\Big)$
$\Rightarrow125=\text{x}^3-\frac{1}{\text{x}^3}-(3\times5)$
$\Rightarrow125=\text{x}^3-\frac{1}{\text{x}^3}-15$
$\Rightarrow125+15=\text{x}^3-\frac{1}{\text{x}^3}$
$\Rightarrow\text{x}^3-\frac{1}{\text{x}^3}=140$
Hence, the result is $\text{x}^3-\frac{1}{\text{x}^3}=140.$
View full question & answer→Question 294 Marks
Find the value of $27x^3 + 8y^3,$ if: $3x + 2y = 20$ and $\text{xy}=\frac{14}{9}$
AnswerGiven $3x + 2y = 20, \text{xy}=\frac{14}{9}$
On cubing both sides we get,
$(3x + 2y)^3 = (20)^3$
We shall use identity$ (a + b)^3 = a^3 + b^3 + 3ab(a + b)$
$27x^3 + 8y^3 + 3(3x)(2y)(3x + 2y) = 20 \times 20 \times 20$
$27x^3 + 8y^3 + 18(xy)(3x + 2y) = 8000$
$27\text{x}^3+8\text{y}^3+18\Big(\frac{14}{9}\Big)(20)=8000$
$27x^3 + 8y^3 + 560 = 8000$
$27x^3 + 8y^3 = 8000 - 560$
$27x^3 + 8y^3 = 7440$
Hence the value of $27x^3 + 8y^3$ is $7440.$
View full question & answer→Question 304 Marks
Find the cube the following binomial expressions: $4-\frac{1}{3\text{x}}$
AnswerGiven,
$4-\frac{1}{3\text{x}}$
The above equation is in the form of $(a - b)^{3 }= a^3 - b^3 - 3ab(a - b)$
We know that, $\text{a} = 4, \text{b} =\frac{1}{3\text{x}}$
By using $(a - b)^{3 }$formula
$\Big(4-\frac{1}{3\text{x}}\Big)^3$
$=4^3-\Big(\frac{1}{3\text{x}}\Big)^3-3(4)\Big(\frac{1}{3\text{x}}\Big)\Big(4-\frac{1}{3\text{x}}\Big)$
$=64-\frac{1}{27\text{x}^3}-\frac{12}{3\text{x}}\Big(4-\frac{1}{3\text{x}}\Big)$
$=64-\frac{1}{27\text{x}^3}-\frac{4}{\text{x}}\Big(4-\frac{1}{3\text{x}}\Big)$
$=64-\frac{1}{27\text{x}^3}-\Big(\frac{4}{3\text{x}}\times4\Big)+\Big(\frac{4}{3\text{x}}\times\frac{1}{3\text{x}}\Big)$
$=64-\frac{1}{27\text{x}^3}-\frac{16}{\text{x}}+\Big(\frac{4}{3\text{x}^2}\Big)$
Hence
The cube of $\Big(4-\frac{1}{3\text{x}}\Big)^3=64-\frac{1}{27\text{x}^3}-\frac{16}{\text{x}}+\Big(\frac{4}{3\text{x}^2}\Big)$
View full question & answer→Question 314 Marks
Simplify the following expressions: $(x^2 - x + 1)^2 - (x^2 + x + 1)^2$
AnswerExpanding, we get
$[x^2 - x + 1]^2 - [x^2 + x + 1]^2$
$= (x^2)^2 + (-x)^2 + 1^2 + 2(x^2)(-x) + 2(-x)(1) + 2x^2) - [(x^2)^2 + x^2 + 1 + 2x^2x + 2x(1) + 2x^2(1)]$
$\big[\therefore(x + y + z)^2 = x^2 + y^2 + z^2 + 2xy + 2yz + 2xz\big]$
$= x^{4 }+ y^2 + 1 - 2x^3 − 2x + 2x^2 - x^2 - x^4 - 1 - 2x^3 - 2x - 2x^2$
$= -4x^3 - 4x$
$= -4x(x^2 + 1)$
Hence simplified equation $= [x^2 - x + 1]^2 - [x^2 + x + 1]^2 $
$= -4x(x^2 + 1)$
View full question & answer→Question 324 Marks
Simplify: $(a + b + c)^2 + (a - b + c)^2$
AnswerIn the given problem, we have to simplify the expressions
Given $(a + b + c)^2 + (a - b + c)^2$
By using identity $(x + y + z)^2 = x^2 + y^2 + z^2 + 2xy + 2yz + 2zx$
Hence the equation becomes
$\big(\text{a}+\text{b}+\text{c}\big)+\big(\text{a}-\text{b}+\text{c}\big)$
$=\Big[\text{a}^2+\text{b}^2+\text{c}^2+2\text{ab}+2\text{bc}+2\text{ca}\Big]$
$+\Big[\text{a}^2+(-\text{b})^2+\text{c}^2+2\text{a}(-\text{b})+2(-\text{b})(\text{c})+2\text{ca}\Big]$
$=\text{a}^2+\text{b}^2+\text{c}^2+2\text{ab}+2\text{bc}+2\text{ca}$
$+\text{a}^2+\text{b}^2+\text{c}^2-2\text{ab}-2\text{bc}+2\text{ca}$
$=\text{a}^2+\text{a}^2+\text{b}^2+\text{b}^2+\text{c}^2+\text{c}^2$
$+2\text{ab}-2\text{ab}+2\text{bc}-2\text{bc}+2\text{ca}+2\text{ca}$
$=2\text{a}^2+2\text{b}^2+2\text{c}^2+4\text{ca}$
Talking 2 as common factor we get
$=2\big(\text{a}^2+\text{b}^2+\text{c}^2+2\text{ca}\big)$
Hence the simplified value of $\big(\text{a}+\text{b}+\text{c}\big)^2+\big(\text{a}-\text{b}+\text{c}\big)^2$ is $2\big(\text{a}^2+\text{b}^2+\text{c}^2+2\text{ca}\big).$
View full question & answer→Question 334 Marks
If $2x + 3y = 8$ and $xy = 2$ , find the value of $4x^2 + 9y^2.$
AnswerIn the given problem, we have to find $4x^2 + 9y^2$
We have been given $2x + 3y = 8$ and $xy = 2$
Let us take $2x + 3y = 8$
On squaring both side we get,
$(2x + 3y)^2 = (8)^2$
We shall use the identity $(x + y)^2 = x^2 + 2xy + y^2$
$(2x \times 2x +3y \times 3y + 2 \times 2x \times 3y) = 64$
$4x^2 + 9y^2 + 12xy = 64$
$4x^2 + 9y^2 + 12(xy) = 64$
By substituting $xy = 2$ we get
$4x^2 + 9y^2 + 12(2) = 64$
$4x^2 + 9y^2 + 24 = 64$
$4x^2 + 9y^2 = 64 - 24$
$4x^2 + 9y^2 = 40$
Hence the value of $4x^2 + 9y^2$ is $40$
View full question & answer→Question 344 Marks
Find the cube of the following binomial expressions: $\frac{3}{\text{x}}-\frac{2}{\text{x}^2}$
AnswerGiven,$\Big(\frac{3}{\text{x}}-\frac{2}{\text{x}^2}\Big)^3$
The above equation is in the form of $(a - b)^{3 }= a^3 - b^3 - 3ab(a - b)$
We know that, $\text{a}=\frac{3}{\text{x}},\text{b}=\frac{2}{\text{x}^2}$
By using $(a - b)^{3 }$formula
$\Big(\frac{3}{\text{x}}-\frac{2}{\text{x}^2}\Big)^3$
$=(3\text{x})^3-\Big(\frac{2}{\text{x}^2}\Big)^3-3\Big(\frac{3}{\text{x}}\Big)\Big(\frac{2}{\text{x}^2}\Big)\Big(\frac{3}{\text{x}}-\frac{2}{\text{x}^2}\Big)$
$=\frac{27}{\text{x}^3}-\frac{8}{\text{x}^6}-3\times\frac{3}{\text{x}}\times\frac{2}{\text{x}^2}\Big(\frac{3}{\text{x}}-\frac{2}{\text{x}^2}\Big)$
$=\frac{27}{\text{x}^3}-\frac{8}{\text{x}^6}-\frac{18}{\text{x}^3}\Big(\frac{3}{\text{x}}-\frac{2}{\text{x}^2}\Big)$
$=\frac{27}{\text{x}^3}-\frac{8}{\text{x}^6}-\Big(\frac{18}{\text{x}^3}\times\frac{3}{\text{x}}\Big)+\Big(\frac{18}{\text{x}^3}\times\frac{2}{\text{x}^2}\Big)$
$=\frac{27}{\text{x}^3}-\frac{8}{\text{x}^6}-\frac{54}{\text{x}^4}+\frac{36}{\text{x}^5}$
Hence $\Big(\frac{3}{\text{x}}-\frac{2}{\text{x}^2}\Big)^3=\frac{27}{\text{x}^3}-\frac{8}{\text{x}^6}-\frac{54}{\text{x}^4}+\frac{36}{\text{x}^5}$
View full question & answer→Question 354 Marks
If $a + b + c = 9$ and $ab + bc + ca = 26,$ find value of $a^3 + b^3 + c^3 - 3abc$.
AnswerWe know that
$a^3 + b^3 + c^3 - 3abc$
$= (a + b + c) (a^2 + b^2 + c^2 - ab - bc - ca)$
$\Rightarrow a^3 + b^2 + c^3 - 3abc$
$ = (a + b + c) [(a^2 + b^2 + c^2) - (ab + bc + ca)] ...(1)$
It follow from the above identity that we require the values of $a + b + c, a^2 + b^2 + c^2,$ and $ab + bc + ca$ to get the value of $a^3 + b^3 + c^3 - 3abc.$
The values of $a + b + c $ and $ab + bc + ca$ are known to us.
So we require the value of $a^2 + b^2 + c^2,$
Now,
$(a + b + c)^2$
$= a^2 + b^2 + c^2 + 2(ab + bc + ca)$
$\Rightarrow (9)^2 = a^2 + b^2 + c^2 + 2 \times 26 [\therefore a + b + c = 9$ and $ab + bc + ca = 26]$
$\Rightarrow 81 = a^2 + b^2 + c^2 + 52$
$\Rightarrow a^2 + b^2 + c^2 = 81 - 52 = 29$
Substituting the values of $a^2 + b^2 + c^{2 }$ in $(1)$, we get,
$a^3 + b^3 + c^3 - 3abc = 9(29 - 26) (\therefore a + b + c = 9 and ab + bc + ca = 26)$
$= 9 \times 3$
$= 27$
$\therefore a^3 + b^3 + c^3 - 3abc = 27$
View full question & answer→Question 364 Marks
If $\text{x}-\frac{1}{\text{x}}=7,$ find the value of $\text{x}^3-\frac{1}{\text{x}^3}.$
AnswerGiven,If $\text{x}-\frac{1}{\text{x}}=7$ We know that, $(a - b)^3 = a^3 - b^3 + 3ab(a + b) ...(1)$
Substitute $\text{x}-\frac{1}{\text{x}}=7$ in $eq(1)$
$\Rightarrow\Big(\text{x}-\frac{1}{\text{x}}\Big)^3=\text{x}^3-\frac{1}{\text{x}^3}-3\Big(\text{x}\times\frac{1}{\text{x}}\Big)\Big(\text{x}-\frac{1}{\text{x}}\Big)$
$\Rightarrow7^3=\text{x}^3-\frac{1}{\text{x}^3}-3\Big(\text{x}\times\frac{1}{\text{x}}\Big)\Big(\text{x}-\frac{1}{\text{x}}\Big)$
$\Rightarrow343=\text{x}^3-\frac{1}{\text{x}^3}-(3\times7)$
$\Rightarrow343=\text{x}^3-\frac{1}{\text{x}^3}-21$
$\Rightarrow125+21=\text{x}^3-\frac{1}{\text{x}^3}$
$\Rightarrow\text{x}^3-\frac{1}{\text{x}^3}=364$
Hence, the result is $\text{x}^3-\frac{1}{\text{x}^3}=364.$
View full question & answer→Question 374 Marks
If $a + b + c = 9$ and $a^2 + b^2 + c^2 = 35$, find value of $a^3 + b^3 + c^3 - 3abc.$
AnswerWe know that
$a^3 + b^3 + c^3 - 3abc$
$= (a + b + c) (a^2 + b^2 + c^2 - ab - bc - ca)$
$\Rightarrow a^3 + b^3 + c^3 - 3abc$
$ = (a + b + c) [(a^2 + b^2 + c^2) - (ab + bc + ca)] ...(1)$
It follow from the above identity that we require the values of $a + b + c, a^2 + b^2 + c^2, $ and $ab + bc + ca$ to get the value of $a^3 + b^3 + c^3 - 3abc.$
The values of $a + b + c$ and $a^2 + b^2 + c^2$ are known to us.
So we require the value of $ab + bc + ca,$
Now,
$(a + b + c)^2 = a^2 + b^2 + c^2 + 2(ab + bc + ca)$
$\Rightarrow (9)^2 = 35 + 2 (ab + bc + ca) [\therefore a + b + c = 9$ and $a^2 + b^2 + c^2 = 35]$
$\Rightarrow 81 = 35 + 2 (ab + bc + ca)$
$\Rightarrow 2(ab + bc + ca) 81 - 35 = 46$
$\Rightarrow\text{ab}+\text{bc}+\text{ca}=\frac{46}{2}=23$
Substituting the values of $ab + bc + ca^{ }$ in $(1),$ we get,
$a^3 + b^3 + c^3 - 3abc = 9(35 - 23) (\therefore a + b + c = 9$ and $a^2 + b^2 + c^2 = 35)$
$= 9 \times 12$
$= 108$
$\therefore a^3 + b^3 + c^3 - 3abc = 108$
View full question & answer→Question 384 Marks
If $x + y + z = 8$ and $xy + yz + zx = 20,$ find value of $x^3 + y^3 + z^3 - 3xyz.$
AnswerWe know that
$x^3 + y^3 + z^3 - 3xyz = (x + y + z) (x^2 + y^2 + z^2 - xy - yz - zx)$
$\Rightarrow x^3 + y^2 + z^3 - 3xyz$
$ = (x + y + z) [(x^2 + y^2 + z^2) - (xy + yz + zx)] ...(1)$
It follow from the above identity that we require the values of $x + y + z, x^2 + y^2 + z^2,$ and $xy + yz + zx$ to get the value of $x^3 + y^3 + z^3 - 3xyz.$
The values of $x + y + z$ and $xy + yz + zx$ are known to us.
So we require the value of $x^2 + y^2 + z^2,$
Now,
$(x + y + z)^2 $
$= x^2 + y^2 + z^2 + 2(xy + yz + zx)$
$\Rightarrow (8)^2$
$ = x^2 + y^2 + z^2 + 2(20) [\therefore x + y + z = 8$ and $xy + yz + zx = 20]$
$\Rightarrow 64 = x^2 + y^2 + z^2 + 40$
$\Rightarrow x^2 + y^2 + z^2 = 64 - 40 = 24$
Substituting the values of $x^2 + y^2 + z^2, x + y + z$ and $xy + yz + zx$ in equation $(1),$
we get, $x^3 + y^3 + z^3 - 3xyz = 8 \times (24 - 20)$
$= 8 \times 4$
$= 32$
$\therefore x^3 + y^3 + z^3 - 3xyz = 32$
View full question & answer→