Question 514 Marks
If $3x + 2y = 9$ and $xy = 3,$ find $: 27x^3 + 8y^3.$
Answer$27x^3 + 8y^3 $
$ = (3x)^3 + (2y)^3= (3x + 2y)^3 − 3.3x . 2y (3x + 2y)$
$= (3x − 2y)^3 − 18xy (3x + 2y)$
$= (9)^3− 18(3) (9)$
$= 729 − 486$
$= 243$
View full question & answer→Question 524 Marks
Evaluate: $(2 − z) (15 − z)$
Answer$(2 − z) (15 − z)$
$ = 2(15 − z) −z(15 − z)= 30 − 2z − 15z + z^2$
$= 30 − 17z + z^2$
View full question & answer→Question 534 Marks
If $a + b = 8$ and $ab = 15,$ find $: a^3 + b^3.$
Answer$a^3 + b^3$
$ = (a + b)^3 − 3ab (a + b)= (8)^3 − 3(15) (8)$
$= 512 − 360$
$= 152$
View full question & answer→Question 544 Marks
Evaluate: $(9 − y) (7 + y)$
Answer$(9 − y) (7 + y)$
$ = 9(7 + y) − y (7 + y)= 63 + 9y − 7y − y^2$
$= 63 + 2y − y^2$
View full question & answer→Question 554 Marks
If $3x – 4y = 5$ and $xy = 3,$ find : $27x^3 – 64y^3.$
Answer$27x^3 – 64x^3 = (3x)^3 – (4y)^3$
$= (3x − 4y)^3 (3x − 4y)^3 + 3 (3x) (4y) (3x − 4y) ......[ \because a^3 − b^3= (a − b)^3 + 3ab(a − b)]$
$= (5)^3 + 36(xy) (3x − 4y)$
$= 125 + 36 (3) (5)$
$= 125 + 540$
$= 665$
View full question & answer→Question 564 Marks
Evaluate : $(2a + 0.5) (7a − 0.3)$
Answer$ (2a + 0.5) (7a − 0.3)$
$= 2a (7a − 0.3) + 0.5 (7a − 0.3)$
$= 14a^2− 0.6a + 3.5a − 0.15$
$= 14a^2 + 2.9a − 0.15$
View full question & answer→Question 574 Marks
If $a + b + c = 11$ and $a^2 + b^2 + c^2 = 81,$ find $: ab + bc + ca.$
AnswerSince $(a+b+c)^2=a^2+b^2+c^2+2(a b+b c+c a)$
$\therefore(11)^2=81+2(a b+b c+c a)$
$\therefore 2(a b+b c+c a)=121-81=40$
$a b+b c+c a=\frac{40}{2}$
$\Rightarrow \mathrm{ab}+\mathrm{bc}+\mathrm{ca}=20$
View full question & answer→Question 584 Marks
Evaluate: $\left(3 \mathrm{x}+\frac{1}{2}\right)\left(2 \mathrm{x}+\frac{1}{3}\right)$
Answer$ \left(3 \mathrm{x}+\frac{1}{2}\right)\left(2 \mathrm{x}+\frac{1}{3}\right)$
$ =3 \mathrm{x}\left(2 \mathrm{x}+\frac{1}{3}\right)+\frac{1}{2}\left(2 \mathrm{x}+\frac{1}{3}\right)$
$ =6 \mathrm{x}^2+\mathrm{x}+\mathrm{x}+\frac{1}{6}$
$ =6 \mathrm{x}^2+2 \mathrm{x}+\frac{1}{6}$
View full question & answer→Question 594 Marks
If $a + b + c = 9$ and $ab + bc + ca = 15,$ find$: a^2 + b^2 + c^2.$
AnswerSince $(a + b + c)^2= a^2 + b^2 + c^2 + 2 (ab + bc + ca)$
$\therefore (9)^2 = a^2 + b^2+ c^2 + 2 (15)$
$81 = a^2+ b^2 + c^2 + 30$
$\therefore a^2 + b^2 + c^2= 81 − 30 = 51$
View full question & answer→Question 604 Marks
If $a + b + c = 10 $ and $a^2 + b^2 + c^2 = 38,$ find $: ab + bc + ca$
Answer$a+b+c=10$
$ \Rightarrow(a+b+c)^2=(10)^2$
$ \Rightarrow a^2+b^2+c^2+2 a b+2 b c+2 c a=100$
$ \Rightarrow 38+2(a b+b c+c a)=100$
$ \Rightarrow 2(a b+b c+c a)=100-38$
$ \Rightarrow 2(a b+b c+c a)=62$
$ \Rightarrow(a b+b c+c a)=\frac{62}{2}$
$ \Rightarrow a b+b c+c a=31$
Alternative Method :
$(a+b+c)^2=a^2+b^2+c^2+2 a b+2 b c+2 c a$
$ \Rightarrow(10)^2=38+2(a b+b c+c a)$
$ \Rightarrow 100=38+2(a b+b c+c a)$
$ \Rightarrow 100-38=2(a b+b c+c a)$
$ \Rightarrow 62=2(a b+b c+c a)$
$ \Rightarrow \frac{62}{2}=a b+b c+c a$
$ \Rightarrow 31=a b+b c+c a$
$\therefore ab + bc + ca = 31$
View full question & answer→Question 614 Marks
If $\mathrm{a}^2+\frac{1}{\mathrm{a}^2}=11$, find : $\mathrm{a}-\frac{1}{\mathrm{a}}$
Answer$ \left(\mathrm{a}-\frac{1}{\mathrm{a}}\right)^2=\mathrm{a}^2+\frac{1}{\mathrm{a}^2}-2$
$ \Rightarrow\left(\mathrm{a}-\frac{1}{\mathrm{a}}\right)^2=11-2$
$ \Rightarrow\left(\mathrm{a}-\frac{1}{\mathrm{a}}\right)^2=9$
$ \Rightarrow \mathrm{a}-\frac{1}{\mathrm{a}}=\sqrt{9}$
$ \Rightarrow \mathrm{a}-\frac{1}{\mathrm{a}}= \pm 3$
View full question & answer→Question 624 Marks
If $\mathrm{a}^2+\frac{1}{\mathrm{a}^2}=23$, find $: \mathrm{a}+\frac{1}{\mathrm{a}}$
Answer$ \left(\mathrm{a}+\frac{1}{\mathrm{a}}\right)^2=\mathrm{a}^2+\frac{1}{\mathrm{a}^2}+2$
$ \Rightarrow\left(\mathrm{a}+\frac{1}{\mathrm{a}}\right)^2=23+2$
$ \Rightarrow\left(\mathrm{a}+\frac{1}{\mathrm{a}}\right)^2=25$
$ \Rightarrow \mathrm{a}+\frac{1}{\mathrm{a}}=\sqrt{25}$
$ \Rightarrow \mathrm{a}+\frac{1}{\mathrm{a}}= \pm 5$
View full question & answer→Question 634 Marks
If $\mathrm{a}-\frac{1}{\mathrm{a}}=4$, find $: \mathrm{a}^2+\frac{1}{\mathrm{a}^2}$
Answer$\left(\mathrm{a}-\frac{1}{\mathrm{a}}\right)^2=\mathrm{a}^2+\frac{1}{\mathrm{a}^2}-2$
$ \Rightarrow(4)^2=\mathrm{a}^2+\frac{1}{\mathrm{a}^2}-2$
$ \Rightarrow 16=\mathrm{a}^2+\frac{1}{\mathrm{a}^2}-2$
$ \Rightarrow 16+2=\mathrm{a}^2+\frac{1}{\mathrm{a}^2}$
$ \Rightarrow 18=\mathrm{a}^2+\frac{1}{\mathrm{a}^2}$
$ \therefore \mathrm{a}^2+\frac{1}{\mathrm{a}^2}=18$
Alternative Method :
$\mathrm{a}-\frac{1}{\mathrm{a}}=4$
$\Rightarrow\left(\mathrm{a}-\frac{1}{\mathrm{a}}\right)^2=(4)^2$
$ \Rightarrow \mathrm{a}^2+\frac{1}{\mathrm{a}^2}-2=16$
$ \Rightarrow \mathrm{a}^2+\frac{1}{\mathrm{a}^2}=16+2$
$ \Rightarrow \mathrm{a}^2+\frac{1}{\mathrm{a}^2}=18$
View full question & answer→Question 644 Marks
If $a+\frac{1}{a}=3$, find $a^2+\frac{1}{a^2}$
Answer$\left(\mathrm{a}+\frac{1}{\mathrm{a}}\right)^2=\mathrm{a}^2+\frac{1}{\mathrm{a}^2}+2$
$ \Rightarrow(3)^2=\mathrm{a}^2+\frac{1}{\mathrm{a}^2}+2$
$ \Rightarrow 9=\mathrm{a}^2+\frac{1}{\mathrm{a}^2}+2$
$ \Rightarrow 9-2=\mathrm{a}^2+\frac{1}{\mathrm{a}^2}$
$ \Rightarrow 7=\mathrm{a}^2+\frac{1}{\mathrm{a}^2}$
$ \therefore \mathrm{a}^2+\frac{1}{\mathrm{a}^2}=7$
Alternative Method :
$a+\frac{1}{a}=3$
$ \Rightarrow\left(a+\frac{1}{a}\right)^2=(3)^2$
$ \Rightarrow a^2+\frac{1}{a^2}+2=9$
$ \Rightarrow a^2+\frac{1}{a^2}=9-2$
$ \Rightarrow a^2+\frac{1}{a^2}=7$
View full question & answer→Question 654 Marks
If $a^2 + b^2= 10$ and $ab = 3$; find : $a + b$
Answer$ (a+b)^2=a^2+b^2+2 a b$
$ \Rightarrow(a+b)^2=10+2 \times 3$
$ \Rightarrow(a+b)^2=10+6$
$ \Rightarrow(a+b)^2=16$
$ \Rightarrow(a+b)=\sqrt{16}$
$ \Rightarrow a+b= \pm 4$
View full question & answer→Question 664 Marks
If $a^2 + b^2= 10$ and $ab = 3;$ find $: a – b$
Answer$ (a-b)^2=a^2+b^2-2 a b$
$ \Rightarrow(a-b)^2=10-2 \times 3$
$ \Rightarrow(a-b)^2=10-6$
$ \Rightarrow(a-b)^2=4$
$ \Rightarrow(a-b)=\sqrt{4}$
$ \Rightarrow a-b= \pm 2$
View full question & answer→Question 674 Marks
If $a^2 + b^2 = 29$ and $ab = 10$, find :$ a − b$
Answer$ (a-b)^2=a^2+b^2-2 a b$
$ \Rightarrow(a-b)^2=29-2 \times 10$
$ \Rightarrow(a-b)^2=29-20$
$ \Rightarrow(a-b)^2=9$
$ \Rightarrow a-b=\sqrt{9}$
$ \Rightarrow a-b= \pm 3$
View full question & answer→Question 684 Marks
If $a^2 + b^2 = 29$ and $ab = 10$, find $: a + b$
Answer$(a+b)^2=a^2+b^2+2 a b$
$\Rightarrow(a+b)^2=29+2 \times 10$
$\Rightarrow(a+b)^2=29+20$
$\Rightarrow(a+b)^2=49$
$\Rightarrow a+b=\sqrt{49}$
$\Rightarrow \mathrm{a}+\mathrm{b}= \pm 7$
View full question & answer→Question 694 Marks
If $a – b = 6$ and $ab = 16,$ find $a^2 + b^2$
Answer$(a − b)^2 = a^2+ b^2 − 2ab$
$\Rightarrow (6)^2= a^2 + b^2 − 2\times 16$
$\Rightarrow 36 = a^2 + b^2 − 32$
$\Rightarrow 36 + 32 = a^2 + b^2$
$\Rightarrow 68 = a^2 + b^2$
$\therefore a^2 + b^2 = 68$
View full question & answer→Question 704 Marks
If $a+b=5$ and $ab = 6$, find $a^2 + b^2$
Answer$(a + b)^2 = a^2+ b^2 + 2ab$
$\Rightarrow (5)^2 = a^2 + b^2+ 2\times 6$
$\Rightarrow 25 = a^2 + b^2 + 12$
$\Rightarrow 25 − 12 = a^2 + b^2$
$\Rightarrow 13 = a^2 + b^2$
$\therefore a^2 + b^2 = 13$
View full question & answer→Question 714 Marks
The sum of the squares of two numbers is $13$ and their product is $6$. Find : $(i)$ the sum of the two numbers.$(ii)$ the difference between them.
AnswerLet $x$ and $y$ be the two numbers, then,
$x^2+y^2=13 \text { and } x y=6$
$(i) (x+y)^2=x^2+y^2+2 x y$
$=13+2 \times 6$
$=13+12$
$=25$
$\therefore x+y= \pm \sqrt{25}= \pm 5$
$(ii) (x-y)^2=x^2+y^2-2 x y$
$=13-12$
$=1$
$\therefore x-y= \pm 1$
View full question & answer→Question 724 Marks
If $3 x+\frac{1}{3 x}=3$, find : $27 x^3+\frac{1}{27 x^3}$
Answer$ 3 \mathrm{x}+\frac{1}{3 \mathrm{x}}=3$
$ \Rightarrow\left(3 \mathrm{x}+\frac{1}{3 \mathrm{x}}\right)^3=(3)^3$
$ \Rightarrow(3 \mathrm{x})^3+\left(\frac{1}{3 \mathrm{x}}\right)^3+3 \times 3 \mathrm{x} \times \frac{1}{3 \mathrm{x}}\left(3 \mathrm{x}+\frac{1}{3 \mathrm{x}}\right)=27$
$ \Rightarrow 27 \mathrm{x}^3+\frac{1}{27 \mathrm{x}^3}+3\left(3 \mathrm{x}+\frac{1}{3 \mathrm{x}}\right)=27$
$ \Rightarrow 27 \mathrm{x}^3+\frac{1}{27 \mathrm{x}^3}+3(3)=27$
$ \Rightarrow 27 \mathrm{x}^3+\frac{1}{27 \mathrm{x}^3}+9=27$
$ \Rightarrow 27 \mathrm{x}^3+\frac{1}{27 \mathrm{x}^3}=27-9$
$ \Rightarrow 27 \mathrm{x}^3+\frac{1}{27 \mathrm{x}^3}=18$
View full question & answer→Question 734 Marks
If $3 x+\frac{1}{3 x}=3$, find: $9 x^2+\frac{1}{9 x^2}$
Answer$ 3 x+\frac{1}{3 x}=3$
$ \Rightarrow\left(3 x+\frac{1}{3 x}\right)^2=(3)^2$
$ \Rightarrow(3 x)^2+\left(\frac{1}{3 x}\right)^2+2 \times 3 x \times \frac{1}{3 x}=9$
$ \Rightarrow 9 x^2+\frac{1}{9 x^2}+2=9$
$ \Rightarrow 9 x^2+\frac{1}{9 x^2}=9-2$
$ \Rightarrow 9 x^2+\frac{1}{9 x^2}=7$
View full question & answer→Question 744 Marks
If $2 x-\frac{1}{2 x}=4$, find : $8 x^3-\frac{1}{8 x^3}$
Answer$ 2 \mathrm{x}-\frac{1}{2 \mathrm{x}}=4$
$ \Rightarrow\left(2 \mathrm{x}-\frac{1}{2 \mathrm{x}}\right)^3=(4)^3$
$ \Rightarrow(2 \mathrm{x})^3-\left(\frac{1}{2 \mathrm{x}}\right)^3-3 \times 2 \mathrm{x} \times \frac{1}{2 \mathrm{x}}\left(2 \mathrm{x}-\frac{1}{2 \mathrm{x}}\right)=64$
$ \Rightarrow 8 \mathrm{x}^3-\frac{1}{8 \mathrm{x}^3}-3(4)=64$
$ \Rightarrow 8 \mathrm{x}^3-\frac{1}{8 \mathrm{x}^3}-12=64$
$ \Rightarrow 8 \mathrm{x}^3-\frac{1}{8 \mathrm{x}^3}=64+12$
$ \Rightarrow 8 \mathrm{x}^3-\frac{1}{8 \mathrm{x}^3}=76$
View full question & answer→Question 754 Marks
If $2 x-\frac{1}{2 x}=4$, find $: 4 x^2+\frac{1}{4 x^2}$
Answer$ 2 \mathrm{x}-\frac{1}{2 \mathrm{x}}=4$
$ \Rightarrow\left(2 \mathrm{x}-\frac{1}{2 \mathrm{x}}\right)^2=(4)^2$
$ \Rightarrow(2 \mathrm{x})^2+\left(\frac{1}{2 \mathrm{x}}\right)^2-2 \times 2 \mathrm{x} \times \frac{1}{2 \mathrm{x}}=16$
$ \Rightarrow 4 \mathrm{x}^2+\frac{1}{4 \mathrm{x}^2}-2=16$
$ \Rightarrow 4 \mathrm{x}^2+\frac{1}{4 \mathrm{x}^2}=16+2$
$ \Rightarrow 4 \mathrm{x}^2+\frac{1}{4 \mathrm{x}^2}=18$
View full question & answer→Question 764 Marks
Find: $\mathrm{a}^3-\frac{1}{\mathrm{a}^3}$, if $\mathrm{a}-\frac{1}{\mathrm{a}}=4$.
Answer$ a-\frac{1}{a}=4$
$ \Rightarrow\left(a-\frac{1}{a}\right)^3=(4)^3$
$ \Rightarrow a^3-\frac{1}{a^3}-3 a \times \frac{1}{a}\left(a-\frac{1}{a}\right)=64$
$ \Rightarrow a^3-\frac{1}{a^3}+3(4)=64 \ldots \ldots . . .\left[\because-\frac{1}{a}=4\right]$
$ \Rightarrow a^3-\frac{1}{a^3}-12=64$
$ \Rightarrow a^3-\frac{1}{a^3}=64+12$
$ \Rightarrow a^3-\frac{1}{a^3}=76$
View full question & answer→Question 774 Marks
Find: $\mathrm{a}^3+\frac{1}{\mathrm{a}^3}$, if $\mathrm{a}+\frac{1}{\mathrm{a}}=5$.
Answer$ \mathrm{a}+\frac{1}{\mathrm{a}}=5$
$ \Rightarrow\left(\mathrm{a}+\frac{1}{\mathrm{a}}\right)^3=(5)^3$
$ \Rightarrow \mathrm{a}^3+\frac{1}{\mathrm{a}^3}+3 \mathrm{a} \times \frac{1}{\mathrm{a}}\left(\mathrm{a}+\frac{1}{\mathrm{a}}\right)=125$
$ \Rightarrow \mathrm{a}^3+\frac{1}{\mathrm{a}^3}+3(5)=125 \ldots \ldots\left[\mathrm{a}+\frac{1}{\mathrm{a}}=5\right]$
$ \Rightarrow \mathrm{a}^3+\frac{1}{\mathrm{a}^3}+15=125$
$ \Rightarrow \mathrm{a}^3+\frac{1}{\mathrm{a}^3}=125-15$
$ \Rightarrow \mathrm{a}^3+\frac{1}{\mathrm{a}^3}=110$
View full question & answer→Question 784 Marks
If $a – b = 3$ and $ab = 10$, find : $a^3 – b^3.$
Answer$a − b = 3$
$\Rightarrow (a − b)^3 = (3)^3$
$\Rightarrow a^3 − b^3 − 3ab (a − b) = 27$
$\Rightarrow a^3 − b^3− 3\times 10 (3) = 27$
$\Rightarrow a^3 − b^3− 90 = 27$
$\Rightarrow a^3 − b^3 = 27 + 90$
$\Rightarrow a^3 − b^3 = 117$
Alternative Method :
$(a − b)^3 = a^3− b^3 − 3ab (a − b)$
$\Rightarrow (3)^3= a^3 − b^3− 3\times 10 (3)$
$\Rightarrow 27 = a^3− b^3− 90$
$\Rightarrow 27 + 90 = a^3 − b^3$
$\Rightarrow 117 = a^3 − b^3$
$\Rightarrow a^3 − b^3 = 117$
View full question & answer→Question 794 Marks
If $a + b = 6$ and $ab=8,$ find $: a^3 + b^3.$
Answer$a + b = 6$
$\Rightarrow (a + b)^3 = (6)^3$
$\Rightarrow a^3+ b^3 + 3ab (a + b) = 216$
$\Rightarrow a^3 + b^3+ 3\times 8 (6) = 216$
$\Rightarrow a^3 + b^3+ 144 = 216$
$\Rightarrow a^3 + b^3 = 216 − 144$
$\Rightarrow a^3 + b^3 = 72$
Alternative method :
$(a + b)^3 = a^3 + b^3 + 3ab (a + b)$
$\Rightarrow (6)^3 = a^3 + b^3 + 3\times 8 (6)$
$\Rightarrow 216 = a^3+ b^3 + 144$
$\Rightarrow 216 − 144 = a^3 + b^3$
$\Rightarrow 72 = a^3+ b^3$
$\Rightarrow a^3+ b^3 = 72$
View full question & answer→Question 804 Marks
Find : $a + b + c,$ if $a^2 + b^2 + c^2 = 83$ and $ab + bc + ca = 71.$
Answer$ (a+b+c)^2=a^2+b^2+c^2+2 a b+2 b c+2 c a$
$ \Rightarrow(a+b+c)^2=82+2(a b+b c+c a)$
$ \Rightarrow(a+b+c)^2=83+2 \times 71$
$ \Rightarrow(a+b+c)^2=83+142$
$ \Rightarrow(a+b+c)^2=225$
$ \Rightarrow a+b+c=\sqrt{225}$
$ \Rightarrow a+b+c= \pm 15$
View full question & answer→Question 814 Marks
Find : $a^2 + b^2 + c^2,$ if $a + b + c = 9$ and $ab + bc + ca = 24$
Answer$a+ b + c = 9$
$\Rightarrow (a + b + c)^2 = (9)2$
$\Rightarrow a^2 + b^2+ c^2 + 2ab + 2bc + 2ca = 81$
$\Rightarrow a^2+ b^2 + c^2+ 2(ab + bc + ca) = 81$
$\Rightarrow a^2 + b^2 + c^2+ 2\times 24 = 81$
$\Rightarrow a^2 + b^2+ c^2 + 48 = 81$
$\Rightarrow a^2 + b^2+ c^2= 81 − 48$
$\Rightarrow a^2 + b^2 + c^2 = 33$
View full question & answer→Question 824 Marks
Find the cube of: $x-\frac{1}{2}$
Answer$ \left(\mathrm{x}-\frac{1}{2}\right)^3$
$ =(\mathrm{x})^3-\left(\frac{1}{2}\right)^3-3 \times \mathrm{x} \times \frac{1}{2}\left(\mathrm{x}-\frac{1}{2}\right)$
$ =\mathrm{x}^3-\frac{1}{8}-\frac{3 \mathrm{x}}{2}\left(\mathrm{x}-\frac{1}{2}\right)$
$ =\mathrm{x}^3-\frac{1}{8}-\frac{3 \mathrm{x}^2}{2}+\frac{3 \mathrm{x}}{4}$
$ =\mathrm{x}^3-\frac{3 \mathrm{x}^2}{2}+\frac{3 \mathrm{x}}{4}-\frac{1}{8}$
View full question & answer→Question 834 Marks
Find the cube of: $2 x+\frac{1}{x}$
Answer$ \left(2 \mathrm{x}+\frac{1}{\mathrm{x}}\right)^3$
$ =(2 \mathrm{x})^3+\left(\frac{1}{\mathrm{x}}\right)^3+3 \times 2 \mathrm{x} \times \frac{1}{\mathrm{x}}\left(2 \mathrm{x}+\frac{1}{\mathrm{x}}\right)$
$ =8 \mathrm{x}^3+\frac{1}{\mathrm{x}^3}+6\left(2 \mathrm{x}+\frac{1}{\mathrm{x}}\right)$
$ =8 \mathrm{x}^3+\frac{1}{\mathrm{x}^3}+12 \mathrm{x}+\frac{6}{\mathrm{x}}$
$ =8 \mathrm{x}^3+12 \mathrm{x}+\frac{6}{\mathrm{x}}+\frac{1}{\mathrm{x}^3}$
View full question & answer→Question 844 Marks
Find the cube of: $3b − 2a$
Answer$(3b − 2a)^3 $
$= (3b)^3 − (2a)^3 − 3\times 3b\times 2a(3b − 2a)= 27b^3 − 8a^3 − 18ab (3b − 2a)$
$= 27b^3− 8a^3 − 54ab^2 + 36a^2b$
$= 27b^3− 54b^2a + 36ba^2 − 8a^3$
View full question & answer→Question 854 Marks
Find the cube of : $2a + 3b$
Answer$(2a + 3b)^3 $
$= (2a)^3 + (3b)^3+ 3\times 2a\times 3b(2a + 3b)= 8a^3 + 27b^3 + 18ab (2a + 3b)$
$= 8a^3+ 27b^3 + 36a^2b + 54ab^2$
$= 8a^3+ 36a^2b + 54ab^2 + 27b^3$
View full question & answer→Question 864 Marks
Find the cube of : $2a − 1$
Answer$(2a − 1)^3 $
$= (2a)^3 − (1)^3 − 3\times 2a\times 1(2a − 1)$
$= 8a^3− 1 − 6a (2a − 1)$
$= 8a^3 − 1 − 12a^2 + 6a$
$= 8a^3 − 12a^2 + 6a − 1$
View full question & answer→Question 874 Marks
Find the cube of : $a + 2$
Answer$(a + 2)^3 $
$= (a)^3+ (2)^3+ 3\times a\times 2(a + 2)$
$= a^3 + 8 + 6a(a+ 2)$
$= a^3+ 8 + 6a^2+ 12a$
$= a^3+ 6a^2 + 12a+ 8$
View full question & answer→Question 884 Marks
Expand : $\left(2 a-\frac{1}{2 a}\right)^3$
Answer$ \left(2 a-\frac{1}{2 a}\right)^3$
$=(2 a)^3-\left(\frac{1}{2 a}\right)^3-3 \times 2 a \times \frac{1}{2 a}\left(2 a-\frac{1}{2 a}\right)$
$ =8 a^3-\frac{1}{8 a^3}-3\left(2 a-\frac{1}{2 a}\right)$
$ =8 a^3-\frac{1}{8 a^3}-6 a+\frac{3}{2 a}$
View full question & answer→Question 894 Marks
Expand : $\left(a+\frac{1}{a}\right)^3$
Answer$ \left(\mathrm{a}+\frac{1}{\mathrm{a}}\right)^3$
$ =\mathrm{a}^3+\left(\frac{1}{\mathrm{a}}\right)^3+3 \times \mathrm{a} \times \frac{1}{\mathrm{a}} \times\left(\mathrm{a}+\frac{1}{\mathrm{a}}\right)$
$ =\mathrm{a}^3+\frac{1}{\mathrm{a}^3}+3\left(\mathrm{a}+\frac{1}{\mathrm{a}}\right)$
$ =\mathrm{a}^3+\frac{1}{\mathrm{a}^3}+3 \mathrm{a}+\frac{3}{\mathrm{a}}$
View full question & answer→Question 904 Marks
Expand : $(x + 5y)^3$
Answer$(x+ 5y)^3$
$= (x)^3+ (5y)^3 + 3\times x \times 5y (x + 5y)= x^3 + 125y^3 + 15xy (x + 5y)$
$= x^3 + 125y^3 + 15x^2y + 75xy^2$
View full question & answer→Question 914 Marks
Expand : $ (3x − 2y)^3$
Answer$(3x − 2y)^2 $
$= (3x)^3− (2y)^3− 3 \times 3x \times 2y (3x − 2y)$
$= 27x^3− 8y^3 − 18xy (3x − 2y)$
$= 27x^3− 8y^3− 54x^2y + 36xy^2$
View full question & answer→Question 924 Marks
Expand : $(a − 2b)^3$
Answer$(a − 2b)^3 $
$= (a)^3− (2b)^3− 3\times a\times 2b (a − 2b) .........[(a − b)^3= a^3− b^3− 3ab(a − b)]= a^3− 8b^3− 6ab (a − 2b)$
$= a^3 − 8b^3 − 6a^2b + 12ab^2$
View full question & answer→Question 934 Marks
Expand : $(2a + b)^3$
Answer$(2a + b)^3 $
$= (2a)^3 + (b)^3 + 3\times 2a\times b(2a + b) ......[(a + b)^3 = a^3+ b^3 + 3ab(a + b)]= 8a^3 + b^3 + 6ab (2a + b)$
$= 8a^3+ b^3+ 12a^2b + 6ab^2$
View full question & answer→Question 944 Marks
Evaluate using the expansion of $(a + b)^2$ or $(a – b)^2: (20.7)^2$
Answer$(20.7)^2 = (20 + 0.7)^2$
$= (20)^2 + (0.7)^2 + 2 (20) (0.7)$
$= 400 + 0.49 + 28$
$= 428 + 0.49$
$= 428.49$
View full question & answer→Question 954 Marks
Evaluate using the expansion of $(a + b)^2$ or $(a – b)^2: (9.4)^2$
Answer$(9.4)^2= (10 − 0.6)^2$
$= (10)^2 + (0.6)^2 − 2 (10) (0.6)$
$= 100 + 0.36 − 12$
$= 88 + 0.36 $
$= 88.36$
View full question & answer→Question 964 Marks
Evaluate using the expansion of $(a + b)^2$ or $(a – b)^2: (188)^2$
Answer$(188)^2 = (200 − 12)^{2 }$
$= (200)^2+ (12)^2 − 2(200) (12)$
$= 40000 + 144 − 4800$
$= 40144 − 4800$
$= 35344$
View full question & answer→Question 974 Marks
Evaluate using the expansion of $(a + b)^2$ or $(a – b)^2 :(415)^2$
Answer$(415)^2 = (400 + 15)^2= (400)^2+ (15)^2+ 2(400)(15)$
$= 160000 + 225 + 12000$
$= 172225$
View full question & answer→Question 984 Marks
Evaluate using the expansion of $(a + b)^2$ or $(a – b)^2 : (92)^2$
Answer$(92)^2 = (100 − 8)^{2 }$
$= (100)^2+ (8)^2 − 2(100) (8)$
$= 10000 + 64 − 1600$
$= 10064 − 1600$
$= 8464$
View full question & answer→Question 994 Marks
Evaluate using the expansion of $(a + b)^2$ or $(a – b)^2 : (208)^2$
Answer$(208)^2 = (200 + 8)^2$
$= (200)^2+ (8)^2 + 2(200) (8) $
$= 40000 + 64 + 3200$
$= 43264$
View full question & answer→Question 1004 Marks
Find the square of $x+\frac{1}{x}-1$
Answer$ \left(\mathrm{x}+\frac{1}{\mathrm{x}}-1\right)^2$
$=(\mathrm{x})^2+\left(\frac{1}{\mathrm{x}}\right)^2+(-1)^2+2 \times \mathrm{x} \times \frac{1}{\mathrm{x}}+2 \times \frac{1}{\mathrm{x}} \times(-1)+2(-1) \times \mathrm{x}$
$ =\mathrm{x}^2+\frac{1}{\mathrm{x}^2}+1+2-\frac{2}{\mathrm{x}}-2 \mathrm{x}$
$ =\mathrm{x}^2+\frac{1}{\mathrm{x}^2}+3-\frac{2}{\mathrm{x}}-2 \mathrm{x}$
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