4 Marks Questions - Maths STD 9 Questions - Vidyadip

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Question 14 Marks

Simplify the following:
(x + 3)³ + (x - 3)³

Answer

In the given problem, we have to simplify equation
Given (x + 3)³ + (x - 3)³
We shall use the identity a³ + b³ = (a + b)(a² + b² - 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 $2\text{x}^2+54\text{x}.$

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Question 24 Marks

Simplify the following:
$\Big(\frac{\text{x}}{2}+\frac{\text{y}}{3}\Big)^3-\Big(\frac{\text{x}}{2}-\frac{\text{y}}{3}\Big)^3$

Answer

Given $\Big(\frac{\text{x}}{2}+\frac{\text{y}}{3}\Big)^3-\Big(\frac{\text{x}}{2}-\frac{\text{y}}{3}\Big)^3$
We shall use the identity a³ - b³ = (a - b)(a² + b²+ 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 $\frac{\text{x}^2\text{y}}{2}+\frac{2\text{y}^3}{27}.$

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Question 34 Marks

Simplify the following products:
$\Big(\text{m}+\frac{\text{n}}{7}\Big)^3\Big(\text{m}-\frac{\text{n}}{7}\Big)$

Answer

In 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 $\Big(\text{m}+\frac{\text{n}}{7}\Big)^2\Big(\text{m}^2-\frac{\text{n}^2}{49}\Big)$

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Question 44 Marks

Simplify the following products:
$\Big(\frac{1}{2}\text{a}-3\text{b}\Big)\Big(3\text{b}+\frac{1}{2}\text{a}\Big)\Big(\frac{1}{4}\text{a}^2+9\text{b}^2\Big)$

Answer

In 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 $\frac{1}{16}\text{a}^4-81\text{b}^4.$

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Question 54 Marks

Simplify the following:
(2x - 5y)³ - (2x + 5y)³

Answer

Given (2x - 5y)³ - (2x + 5y)³
We shall use the identity a³ - b³ = (a - b)(a² + b² + 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 $-120\text{x}^2\text{y}-250\text{y}^3.$

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Question 64 Marks

Simplify the following expressions:
(x² - x + 1)² - (x² + x + 1)²

Answer

Expanding, we get
[x² - x + 1]² - [x² + x + 1]²
= (x²)² + (-x)² + 1² + 2(x²)(-x) + 2(-x)(1) + 2x²) - [(x²)² + x² + 1 + 2x²x + 2x(1) + 2x²(1)]
$\big[\therefore$(x + y + z)² = x² + y² + z² + 2xy + 2yz + 2xz$\big]$
= x⁴+ y² + 1 - 2x³ − 2x + 2x² - x² - x⁴ - 1 - 2x³ - 2x - 2x²
= -4x³ - 4x
= -4x(x² + 1)
Hence simplified equation = [x² - x + 1]² - [x² + x + 1]² = -4x(x² + 1)

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Question 74 Marks

Simplify:
(x² + y² - z²)² - (x² - y² + z²)²

Answer

We have (x² + y² - z²)² - (x² - y² + z²)²
Using formula (x + y + z)² = x² + y² + z² + 2xy + 2yz + 2zx, we get
(x² + y² - z²)² - (x² - y² + z²)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 4x² 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 $4\text{x}^2(\text{y}^2-\text{z}^2).$

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Question 84 Marks

Simplify:
(a + b + c)² + (a - b + c)²

Answer

In 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 = x² + y² + z² + 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 $2\big(\text{a}^2+\text{b}^2+\text{c}^2+2\text{ca}\big).$

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Question 94 Marks

Simplify:
(a + b + c)²- (a - b + c)²

Answer

In 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 = x² + y² + z² + 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 $4\big(\text{ab}+\text{bc}\big).$

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Question 104 Marks

Simplify:
(2x + p - c)² - (2x - p + c)²

Answer

Given (2x + p - c)² - (2x - p + c)²
By using identity (x + y + z) = x² + y² + z² + 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 $8\text{x}(\text{p}-\text{c}).$

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Question 114 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.

Answer

In 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,
$\text{a}^2+\text{b}^2+\text{c}^2-\text{ab}-\text{bc}-\text{ca}$
$\text{a}^2+\text{b}^2+\text{c}^2-\text{ab}-\text{bc}-\text{ca}\\=\frac{1}{2}\big(2\text{a}^2+2\text{b}^2+2\text{c}^2-2\text{ab}-2\text{bc}-2\text{ca}\big)$
$=\frac{1}{2}[(\text{a}-\text{b}^2)+(\text{b}-\text{c}^2)+(\text{c}-\text{a}^2)]$

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Question 124 Marks

If x + y + z = 8 and xy + yz + zx = 20, find value of x³ + y³ + z³ - 3xyz.

Answer

We know that

x³ + y³ + z³ - 3xyz = (x + y + z) (x² + y² + z² - xy - yz - zx)

⇒ x³ + y² + z³ - 3xyz = (x + y + z) [(x² + y² + z²) - (xy + yz + zx)] ...(1)

It follow from the above identity that we require the values of x + y + z, x² + y² + z², and xy + yz + zx to get the value of x³ + y³ + z³ - 3xyz.

The values of x + y + z and xy + yz + zx are known to us.

So we require the value of x² + y² + z²,

Now,

(x + y + z)² = x² + y² + z² + 2(xy + yz + zx)

⇒ (8)² = x² + y² + z² + 2(20) $[\therefore$ x + y + z = 8 and xy + yz + zx = 20$]$

⇒ 64 = x² + y² + z² + 40

⇒ x² + y² + z² = 64 - 40 = 24

Substituting the values of x² + y² + z², x + y + z and xy + yz + zx in equation (1),

we get, x³ + y³ + z³ - 3xyz = 8 × (24 - 20)

= 8 × 4

= 32

$\therefore$ x³ + y³ + z³ - 3xyz = 32

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Question 134 Marks

If $\text{x}+\frac{1}{\text{x}}=5,$ find the value of $\text{x}^3+\frac{1}{\text{x}^3}.$

Answer

Given, $\text{x}+\frac{1}{\text{x}}=5$
We know that, (a + b)³ = a³ + b³ + 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 $\text{x}^3+\frac{1}{\text{x}^3}=110.$

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Question 144 Marks

If $\text{x}-\frac{1}{\text{x}}=\frac{1}{2},$ then write the value of $4\text{x}^2+\frac{4}{\text{x}^2}.$

Answer

We 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 = a² - 2ab + b²
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.

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Question 154 Marks

If $\text{x}-\frac{1}{\text{x}}=7,$ find the value of $\text{x}^3-\frac{1}{\text{x}^3}.$

Answer

Given,If $\text{x}-\frac{1}{\text{x}}=7$

We know that, (a - b)³ = a³ - b³ + 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 $\text{x}^3-\frac{1}{\text{x}^3}=364.$

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Question 164 Marks

If $\text{x}-\frac{1}{\text{x}}=5,$ find the value of $\text{x}^3-\frac{1}{\text{x}^3}.$

Answer

Given,If $\text{x}-\frac{1}{\text{x}}=5$
We know that, (a - b)³ = a³ - b³ - 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 $\text{x}^3-\frac{1}{\text{x}^3}=140.$

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Question 174 Marks

If $\text{x}+\frac{1}{\text{x}}=3,$ then find the value of $\text{x}^2+\frac{1}{\text{x}^2}.$

Answer

We 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 = a² + 2ab + b²
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.

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Question 184 Marks

If $\text{x}-\frac{1}{\text{x}}=3+2\sqrt2,$ find the value of $\text{x}^3-\frac{1}{\text{x}^3}.$

Answer

In 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 = a³ + b³ + 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 $108+76\sqrt2.$

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Question 194 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 $\text{x}+\frac{1}{\text{x}}.$

Answer

In 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 = a³ + b³ + 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.

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Question 204 Marks

If $\text{x}^2+\frac{1}{\text{x}^2}=98,$ find the value of $\text{x}^3+\frac{1}{\text{x}^3}.$

Answer

Given, $\text{x}^2+\frac{1}{\text{x}^2}=98$
We know that, (x + y)² = x² + y² + 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, a³ + b³ = (a - b)(a² + b² - 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 $\text{x}^3+\frac{1}{\text{x}^3}=970.$

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Question 214 Marks

If $\text{x}^2+\frac{1}{\text{x}^2}=79,$ find the value of $\text{x}+\frac{1}{\text{x}}.$

Answer

In 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 $\pm9$

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Question 224 Marks

If $\text{x}^2+\frac{1}{\text{x}^2}=51,$ find the value of $\text{x}^3-\frac{1}{\text{x}^3}.$

Answer

Given, $\text{x}^2+\frac{1}{\text{x}^2}=51$
We know that, (x - y)² = x² + y² - 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, a³ - b³ = (a - b)(a² + b² + 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 $\text{x}^3-\frac{1}{\text{x}^3}=364.$

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Question 234 Marks

If $\text{a}^2-\frac{1}{\text{a}^2}=102,$ find the value of $\text{a}-\frac{1}{\text{a}}.$

Answer

We 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 = x² + y² - 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.

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Question 244 Marks

If $\text{x}+\frac{1}{\text{x}}=\sqrt{5}$ find the value of $\text{x}^2+\frac{1}{\text{x}^2}$ and $\text{x}^4+\frac{1}{\text{x}^4}$

Answer

We 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, $\text{x}^2+\frac{1}{\text{x}^2}=3,\ \text{x}^4+\frac{1}{\text{x}^4}=7.$

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Question 254 Marks

If a - b = 5 and ab = 12, find the value of a² + b².

Answer

We have to find the value a² + b²
Given a - b = 5, ab = 12
Using identity (a - b)² = a² - 2ab + b²
By substituting the value of a - b = 5, ab = 12 we get ,
(5)² = a² + b² - 2 × 12
5 × 5 = a² + b² - 2 × 12
25 = a² + b²- 24
By transposing -24 to left hand side we get
25 + 24 = a² + b²
49 = a² + b²
Hence the value of a² + b² is 49.

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Question 264 Marks

If a + b + c = 9 and a² + b² + c² = 35, find value of a³ + b³ + c³ - 3abc.

Answer

We know that

a³ + b³ + c³ - 3abc = (a + b + c) (a² + b² + c² - ab - bc - ca)

⇒ a³ + b³ + c³ - 3abc = (a + b + c) [(a² + b² + c²) - (ab + bc + ca)] ...(1)

It follow from the above identity that we require the values of a + b + c, a² + b² + c², and ab + bc + ca to get the value of a³ + b³ + c³ - 3abc.

The values of a + b + c and a² + b² + c² are known to us.

So we require the value of ab + bc + ca,

Now,

(a + b + c)² = a² + b² + c² + 2(ab + bc + ca)

⇒ (9)² = 35 + 2 (ab + bc + ca) $[\therefore$ a + b + c = 9 and a² + b² + c² = 35$]$

⇒ 81 = 35 + 2 (ab + bc + ca)

⇒ 2(ab + bc + ca) 81 - 35 = 46

$\Rightarrow\text{ab}+\text{bc}+\text{ca}=\frac{46}{2}=23$

Substituting the values of ab + bc + cain (1), we get,

a³ + b³ + c³ - 3abc = 9(35 - 23) $(\therefore$ a + b + c = 9 and a² + b² + c² = 35$)$

= 9 × 12

= 108

$\therefore$ a³ + b³ + c³ - 3abc = 108

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Question 274 Marks

If a + b + c = 9 and ab + bc + ca = 26, find value of a³ + b³ + c³ - 3abc.

Answer

We know that

a³ + b³ + c³ - 3abc = (a + b + c) (a² + b² + c² - ab - bc - ca)

⇒ a³ + b² + c³ - 3abc = (a + b + c) [(a² + b² + c²) - (ab + bc + ca)] ...(1)

It follow from the above identity that we require the values of a + b + c, a² + b² + c², and ab + bc + ca to get the value of a³ + b³ + c³ - 3abc.

The values of a + b + c and ab + bc + ca are known to us.

So we require the value of a² + b² + c²,

Now,

(a + b + c)² = a² + b² + c² + 2(ab + bc + ca)

⇒ (9)² = a² + b² + c² + 2 × 26 $[\therefore$ a + b + c = 9 and ab + bc + ca = 26$]$

⇒ 81 = a² + b² + c² + 52

⇒ a² + b² + c² = 81 - 52 = 29

Substituting the values of a² + b² + c²in (1), we get,

a³ + b³ + c³ - 3abc = 9(29 - 26) $(\therefore$ a + b + c = 9 and ab + bc + ca = 26$)$

= 9 × 3

= 27

$\therefore$ a³ + b³ + c³ - 3abc = 27

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Question 284 Marks

If a + b = 10 and ab = 16, find the value of a² - ab + b² and a² + ab + b².

Answer

We have,
a² - ab + b² = a²+ b² - ab
= a² + b² - ab + 2ab - 2ab [Adding and substracting 2ab]
= (a² + b² + 2ab) - 3ab $\big[\because$ (a + b)2 = a2 + b2 - 2ab$\big]$
= (a + b)² - 3 × 16 $\big[\because$ a + b = 10 and b = 16$\big]$
= (10)² - 3 × 16
= 100 - 48
= 52
⇒ a² - ab + b² = 52
We have,
a² + ab + b² = a² + ab + b² + ab - ab [Adding and substracting ab]
= (a² + b² + 2ab) - ab
= (a + b)² - ab $\big[\because$ (a + b)2 = a2 + b2 - 2ab$\big]$
= (10)² - 16
= 100 - 16
⇒ a² + ab + b² = 84
Hence, a² - ab + b² = 52, and a² + ab + b² = 84.

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Question 294 Marks

If 3x - 7y = 10 and xy = -1, find the value of 9x² + 49y².

Answer

In the given problem, we have to find 9x² + 49y²
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 × 3x + 7y × 7y - 2 × 3x × 7y) = 100
We shall use the identity (x - y)² = x² + 2xy + y²
9x² + 49y² - 42xy = 100
9x² + 49y² - 42(xy) = 100
By substituting xy = -1 we get
9x² + 49y² - 42(-1) = 100
9x² + 49y² + 42 = 100
9x² + 49y² = 100 - 42
9x² + 49y² = 58
Hence the value of 9x² + 49y² is 58.

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Question 304 Marks

If 2x + 3y = 8 and xy = 2, find the value of 4x² + 9y².

Answer

In the given problem, we have to find 4x² + 9y²
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 = x² + 2xy + y²
(2x × 2x +3y × 3y + 2 × 2x × 3y) = 64
4x² + 9y² + 12xy = 64
4x² + 9y² + 12(xy) = 64
By substituting xy = 2 we get
4x² + 9y² + 12(2) = 64
4x² + 9y² + 24 = 64
4x² + 9y² = 64 - 24
4x² + 9y² = 40
Hence the value of 4x² + 9y² is 40

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Question 314 Marks

If 2x + 3y = 13 and xy = 6, find the value of 8x³ + 27y³.

Answer

Given, 2x + 3y = 13, xy = 6
We know that,
(2x + 3y)³ = 13²
⇒ 8x³ + 27y³ + 3(2x)(3y)(2x + 3y) = 2197
⇒ 8x³ + 27y³ + 18xy(2x + 3y) = 2197
Substitute 2x + 3y = 13, xy = 6
⇒ 8x³ + 27y³ + 18(6)(13) = 2197
⇒ 8x³ + 27y³ + 1404 = 2197
⇒ 8x³ + 27y³ = 2197 – 1404
⇒ 8x³ + 27y³ = 793
Hence, the value of 8x³ + 27y³ = 793

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Question 324 Marks

Find the value of 64x³ - 125z³, if 4x - 5z = 16 and xz = 12.

Answer

Given, 64x³ - 125z³
Here, 4x - 5z = 16 and xz = 12
Cubing 4x - 5z = 16 on both sides
(4x - 5z)³ = 16³
We know that, (a - b)³ = a³ - b³ - 3ab(a - b)
(4x)³ - (5z)³ - 3(4x)(5z)(4x - 5z) = 16³
64x³ - 125z³ - 60(xz)(16) = 4096
64x³ - 125z³ - 60(12)(16) = 4096
64x³ - 125z³ - 11520 = 4096
64x³ - 125z³ = 4096 + 11520
64x³ - 125z³ = 15616
The value of 64x³ - 125z³ = 15616

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Question 334 Marks

Find the value of 27x³ + 8y³, if:
3x + 2y = 20 and $\text{xy}=\frac{14}{9}$

Answer

Given 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 = a³ + b³ + 3ab(a + b)
27x³ + 8y³ + 3(3x)(2y)(3x + 2y) = 20 × 20 × 20
27x³ + 8y³ + 18(xy)(3x + 2y) = 8000
$27\text{x}^3+8\text{y}^3+18\Big(\frac{14}{9}\Big)(20)=8000$
27x³ + 8y³ + 560 = 8000
27x³ + 8y³ = 8000 - 560
27x³ + 8y³ = 7440
Hence the value of 27x³ + 8y³ is 7440.

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Question 344 Marks

Find the value of 27x³ + 8y³, if
3x + 2y = 14 and xy = 8

Answer

In the given problem, we have to find the value of 27x³ + 8y³
Given 3x + 2y = 14, xy = 8
On cubing both sides we get,
(3x + 2y)³ = (14)³
We shall use identity (a + b)³ = a³ + b³+ 3ab(a + b)
27x³ + 8y³ + 3(3x)(2y)(3x + 2y) = 14 × 14 × 14
27x³ + 8y³ + 18(xy)(3x + 2y) = 14 × 14 × 14
27x³ + 8y³ + 18(8)(14) = 2744
27x³ + 8y³ + 2016 = 2744
27x³ + 8y³ = 2744 - 2016
27x³ + 8y³= 728
Hence the value of 27x³ + 8y³ is 728.

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Question 354 Marks

Find the cube the following binomial expressions:
$4-\frac{1}{3\text{x}}$

Answer

Given,
$4-\frac{1}{3\text{x}}$
The above equation is in the form of (a - b)³= a³ - b³ - 3ab(a - b)
We know that, $\text{a} = 4, \text{b} =\frac{1}{3\text{x}}$
By using (a - b)³formula
$\Big(4-\frac{1}{3\text{x}}\Big)^3$
$=4^3-\Big(\frac{1}{3\text{x}}\Big)^3-3(4)\Big(\frac{1}{3\text{x}}\Big)\Big(4-\frac{1}{3\text{x}}\Big)$
$=64-\frac{1}{27\text{x}^3}-\frac{12}{3\text{x}}\Big(4-\frac{1}{3\text{x}}\Big)$
$=64-\frac{1}{27\text{x}^3}-\frac{4}{\text{x}}\Big(4-\frac{1}{3\text{x}}\Big)$
$=64-\frac{1}{27\text{x}^3}-\Big(\frac{4}{3\text{x}}\times4\Big)+\Big(\frac{4}{3\text{x}}\times\frac{1}{3\text{x}}\Big)$
$=64-\frac{1}{27\text{x}^3}-\frac{16}{\text{x}}+\Big(\frac{4}{3\text{x}^2}\Big)$
Hence
The cube of $\Big(4-\frac{1}{3\text{x}}\Big)^3=64-\frac{1}{27\text{x}^3}-\frac{16}{\text{x}}+\Big(\frac{4}{3\text{x}^2}\Big)$

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Question 364 Marks

Find the cube of the following binomial expressions:
$\frac{3}{\text{x}}-\frac{2}{\text{x}^2}$

Answer

Given,

$\Big(\frac{3}{\text{x}}-\frac{2}{\text{x}^2}\Big)^3$

The above equation is in the form of (a - b)³= a³ - b³ - 3ab(a - b)

We know that, $\text{a}=\frac{3}{\text{x}},\text{b}=\frac{2}{\text{x}^2}$

By using (a - b)³formula

$\Big(\frac{3}{\text{x}}-\frac{2}{\text{x}^2}\Big)^3$

$=(3\text{x})^3-\Big(\frac{2}{\text{x}^2}\Big)^3-3\Big(\frac{3}{\text{x}}\Big)\Big(\frac{2}{\text{x}^2}\Big)\Big(\frac{3}{\text{x}}-\frac{2}{\text{x}^2}\Big)$

$=\frac{27}{\text{x}^3}-\frac{8}{\text{x}^6}-3\times\frac{3}{\text{x}}\times\frac{2}{\text{x}^2}\Big(\frac{3}{\text{x}}-\frac{2}{\text{x}^2}\Big)$

$=\frac{27}{\text{x}^3}-\frac{8}{\text{x}^6}-\frac{18}{\text{x}^3}\Big(\frac{3}{\text{x}}-\frac{2}{\text{x}^2}\Big)$

$=\frac{27}{\text{x}^3}-\frac{8}{\text{x}^6}-\Big(\frac{18}{\text{x}^3}\times\frac{3}{\text{x}}\Big)+\Big(\frac{18}{\text{x}^3}\times\frac{2}{\text{x}^2}\Big)$

$=\frac{27}{\text{x}^3}-\frac{8}{\text{x}^6}-\frac{54}{\text{x}^4}+\frac{36}{\text{x}^5}$

Hence $\Big(\frac{3}{\text{x}}-\frac{2}{\text{x}^2}\Big)^3=\frac{27}{\text{x}^3}-\frac{8}{\text{x}^6}-\frac{54}{\text{x}^4}+\frac{36}{\text{x}^5}$

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Question 374 Marks

Find the cube of the following binomial expressions:
$\frac{1}{\text{x}}+\frac{\text{y}}{3}$

Answer

Given,
$\frac{1}{\text{x}}+\frac{\text{y}}{3}$
The above equation is in the form of (a + b)³= a³ + b³ + 3ab(a + b)
We know that,$ \text{a} =\frac{1}{\text{x}},\text{b} =\frac{\text{y}}{3}$
By using (a + b)³formula
$\Big(\frac{1}{\text{x}}+\frac{\text{y}}{3}\Big)^3$
$=\Big(\frac{1}{\text{x}}\Big)^3+\Big(\frac{\text{y}}{3}\Big)^3+3\Big(\frac{1}{\text{x}}\Big)\Big(\frac{\text{y}}{3}\Big)\Big(\frac{1}{\text{x}}+\frac{\text{y}}{3}\Big)$
$=\frac{1}{\text{x}^3}+\frac{\text{y}^3}{27}+3\times\frac{1}{\text{x}}\times\frac{\text{y}}{3}\Big(\frac{1}{\text{x}}+\frac{\text{y}}{3}\Big)$
$=\frac{1}{\text{x}^3}+\frac{\text{y}^3}{27}+\frac{\text{y}}{\text{x}}\Big(\frac{1}{\text{x}}+\frac{\text{y}}{3}\Big)$
$=\frac{1}{\text{x}^3}+\frac{\text{y}^3}{27}+\Big(\frac{\text{y}}{\text{x}}\times\frac{1}{\text{x}}\Big)+\Big(\frac{\text{y}}{\text{x}}\times\text{y}^3\Big)$
$=\Big(\frac{1}{\text{x}^3}+\frac{\text{y}^3}{27}+\frac{\text{y}}{\text{x}^2}+\frac{\text{y}^2}{3\text{x}}\Big)$
Hence
$\Big(\frac{1}{\text{x}}+\frac{\text{y}}{3}\Big)^3=\Big(\frac{1}{\text{x}^3}+\frac{\text{y}^3}{27}+\frac{\text{y}}{\text{x}^2}+\frac{\text{y}^2}{3\text{x}}\Big)$

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Question 384 Marks

Find the cube of the following binomial expressions:
$2\text{x}+\frac{3}{\text{x}}$

Answer

Given,
$2\text{x}+\frac{3}{\text{x}}$
The above equation is in the form of (a + b)³= a³ + b³ + 3ab(a + b)
we know that $\text{a}=2\text{x},\text{b}=\frac{3}{\text{x}}$
By using (a + b)³formula
$=8\text{x}^3+\frac{27}{\text{x}^3}+\frac{18\text{x}}{\text{x}}\Big(\frac{2}{\text{x}}+\frac{3}{\text{x}}\Big)$
$=8\text{x}^3+\frac{27}{\text{x}^3}+\frac{18\text{x}}{\text{x}}\Big(2{\text{x}}+\frac{3}{\text{x}}\Big)$
$=8\text{x}^3+\frac{27}{\text{x}^3}+\Big({18}\times2{\text{x}}\Big)+\Big(18\times\frac{3}{\text{x}}\Big)$
$=\Big(8\text{x}^3+\frac{27}{\text{x}^3}+36\times\frac{54}{\text{x}}\Big)$
Hence
The cube of $\Big(2\text{x}+\frac{3}{\text{x}}\Big)^3=\Big(8\text{x}^3+\frac{27}{\text{x}^3}+36\times\frac{54}{\text{x}}\Big)$

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