ML Aggarwal Solution Class 9 Chapter 3 Expansions Exercise 3.2

Exercise 3.2

Question 1

Sol :

Given : x-y=8..(i)

xy=5...(ii)

Squaring both sides in equation (i)

∴(x-y)²=8²

⇒x²-2.xy+y²=64

⇒x²-2(5)+y²=64

⇒x²+y²=64+10

⇒x²+y²=74

Question 2

Sol :

⇒Given : x+y=10...(i)

⇒xy=21....(ii)

Squaring both sides in equation (i)

⇒(x+y)²=10²

⇒x²+2xy+y²=100

⇒x²+2(21)+y²=100

⇒x²+4²+y²=100

⇒x²+y²=100-42

⇒x²+y²=58

2(x²+y²)=2×58=116

Question 3

Sol :

⇒Given : 2a+3b=7..(i)

⇒ab=2....(ii)

Squaring both sides in equal (i)

⇒(2a+3b)²=7²

⇒(2a)²+2.2a+3b+(3b)²=49

⇒4a²+12ab+9b²=49

⇒4a²+12(2)+9b²=49

⇒4a²+24+9b²=49

⇒4a²+9b²=49-24

⇒4a²+9b²=25

Question 4

Sol :

⇒Given : 3x-4y=16...(i)

⇒xy=4..(ii)

Squaring on both sides in equation (i)

⇒(3x-4y)²=16²

⇒(3x)²-2.3x.4y+(4y)²=256

⇒9x²-24xy+16y²=256

⇒9x²-24(4)+16y²=256

⇒9x²-96+16y²=256

⇒9x²+16y²=256+96

⇒9x²+16y²=352

Question 5

Sol :

⇒Given : x+y=8...(i)

⇒x-y=2...(ii)

Since we know that

⇒2(x)²+2(y)²=(x+y)²+(x-y)²

From (i) ⇒Squaring on both sides

(x+y)²=8²

=64

From (ii) ⇒Squaring on both sides

(x-y)²=2²

=4

∴2(x²+y²)=(x+y)²+(x-y)²

=64+4

=68

Question 6

Sol :

Given : a²+b²=13...(i)

ab=6...(ii)

(i) a+b

∴(a+b)²=a²+2ab+b²

=a²+b²+2ab

=13+2(6)

=13+12

(a+b)²=25

a+b=√25

a+b=5

(ii) a-b

∴(a-b)²=a²-2ab+b²

=a²+b²-2ab

=13-2(6)

=13-12

(a-b)²=1

(a-b)=√1

∴a-b=1

Question 7

Sol :

Given : a+b=4...(i)

ab=-12...(ii)

(i) a-b

from (i) 

⇒ab=-12

⇒a(4-a)=-12

⇒4a-a²=-12

⇒a²-12-4a=0

⇒a²-4a-12=0

⇒a²-6a+2a-12=0

⇒a(a-6)+2(a-6)=0

⇒(a-6)(a+2)=0

∴a-6=0 ; a+2=0

∴a=6 ; a=-2

∴a+b=4 ; a+b=4

6+b=4 ; -2+b=4

b=4-6=-2 ; b=4+2=6

∴(a,b)=(6,-2)

(i) a-b=6-(-2)=8

(ii) a²-b²=(a+b)(a-b)

=(4)(8)

=32

Question 8

Sol :

Given : p-q=9..(i)

pq=36...(ii)

From (i) and (ii)

p=q+9

∴pq=36

⇒(q+9)q=36

⇒q²+9q=36

⇒q²+9q-36=0

⇒q²+19q-3q-36=0

⇒q(q+12)-3(q+12)=0

⇒(q+12)(q-3)=0

∴q+12=0 ; q-3=0

q=-12 ; q=3

∵p-q=9 ; p-q=9

p-(-12)=9 ; p-3=9

p=9-12 ; p=9+3

p=-3 ; p=12

∴p=12 ; q=3

(i) p+q

=12+3=15

(ii) p²-q²

=(p+q)(p-q)

=15.9

=135

Question 9

Sol :

Given : x+y=6..(i) 

x-y=4..(ii)

From (i) ⇒Squaring on both sides

⇒(x+y)²=6²

⇒x²+y²+2xy=36..(i)

⇒x²+y²=36-2xy..(ii)

From (ii) ⇒Squaring on both sides

⇒(x+y)²=4²

⇒x²-2xy+y²=16

⇒x²+y²=16+2xy...(iv)

∴Equate equation (iii) and (iv)

⇒36-2xy=16+2xy

⇒36-16=2xy+2xy

⇒20=4xy

⇒4xy=20

⇒$xy=\frac{20}{4}$

⇒xy=5

(i) x²+y²

⇒From equation (iii) 

x²+y²=36-2xy

=36-2(5)

=36-10=26

(ii) xy=5

Question 10

Sol :

Given : $x-3=\frac{1}{x}$

⇒$x-\frac{1}{x}=3$

Squaring on both sides

⇒$\left(x-\frac{1}{x}\right)^2=3^2$

⇒$x^2-2.x\frac{1}{x}+\frac{1}{x^2}=9$

⇒$x^2-2+\frac{1}{x^2}=9$

⇒$x^2+\frac{1}{x^2}=9+2$

⇒$x^2+\frac{1}{x^2}=11$

Question 11

Sol :

Given : x+y=8 ...(1)

$xy=3\frac{3}{4}=\frac{15}{4}$....(2)

From eq(1) 

⇒y=8-x

From eq(2)

⇒$xy=\frac{15}{4}$

⇒$x(8-x)=\frac{15}{4}$

⇒$8x-x^2=\frac{15}{4}$

⇒4(8x-x²)=15

⇒32x-4x²=15

⇒4x²-32x+15=0

⇒4x²-32x+15=0

⇒4x²-30x-2x+15=0

⇒2x(2x-15)-1(2x-15)=0

⇒(2x-15)(2x-1)=0

∴2x-15=0 ; 2x-1=0

2x=15 ; 2x=1

$x=\frac{15}{2}$ ; $x=\frac{1}{2}$

∴x+y=8 ; x+y=8

$\frac{15}{2}+y=8$  ; $\frac{1}{2}+y=8$

$y=8-\frac{15}{2}$ ; $y=8-\frac{1}{2}$

$y=\frac{16-15}{2}$ ; $y=\frac{16-1}{2}$

$y=\frac{1}{2}$ ; $y=\frac{15}{2}$

∴$x=\frac{15}{2}$  ; $y=\frac{1}{2}$

(i) x-y

⇒$\frac{15}{2}-\frac{1}{2}$

⇒$\frac{15-1}{2}$

⇒$\frac{14}{2}$

⇒7

(ii) 3(x²-y²)

⇒$3\left[\left(\frac{15}{2}\right)^2+\left(\frac{1}{2}\right)^2\right]$

⇒$3\left[\frac{225}{4}+\frac{1}{4}\right]$

⇒$3\left[\frac{225+1}{4}\right]$

⇒$3\left(\frac{226}{4}\right)$

⇒$\frac{3\times 113}{2}$

⇒$\frac{339}{2}$

(iii) 5(x²+y²)+4(x-y)

⇒$5\left(\left(\frac{15}{2}\right)^2+\left(\frac{1}{2}\right)^2\right)+4(7)$

⇒$5\left[\frac{225}{4}+\frac{1}{4}\right]+4(7)$

⇒$5\left(\frac{113}{2}\right)+28$

⇒$\frac{565}{2}+28$

⇒$\frac{565+56}{2}$

⇒$\frac{621}{2}$

Question 12

Sol :

Given : x²+y²=34

$xy=10\frac{1}{2}=\frac{21}{2}$

∴(x+y)²=x²+y²+2xy

$=34+2\times \frac{21}{2}$

=34+21=55

∴(x+y)²=x²+y²-2xy

$=34-2\times \frac{21}{2}$

=34-21=13

∴2(x+y)²+(x-y)²

⇒2(55)+13

⇒110+13

⇒123

Question 13

Sol :

Given : a-b=3...(i)

ab=4...(ii)

∴a³-b³=(a-b)(a²+ab+b²)

From (i) Squaring on both sides

⇒(a-b)²=3²

⇒a²+b²-2ab=9

⇒a²+b²-2(4)=9

⇒a²+b²-8=9

⇒a²+b²=9+8

⇒a²+b²=17

∴a³-b³=(a-b)(a²+ab+b²)

=3(a²+ab+b²)

=3(17+4)

=3(21)=63

Question 14

Sol :

Given : 2a-3b=3...(i)

ab=2..(ii)

From (i) Squaring on both sides

⇒(2a-3b)²=3²

⇒(2a)²-2.2a.3b+(3b)²=9

⇒4a²-12ab+9b²=9

⇒4a²+9b²-12(2)=9

⇒4a²+9b²-24=9

⇒4a²+9b²=9+24

=33

∴(2a)³-(3b)³⇒8a³-27b³

⇒(2a-3b)((2a)²+2a.3b+(3b)²)

⇒3(4a²+6ab+9b²)

⇒3(33+6(2))

⇒3(33+12)

⇒3(45)

⇒139

Question 15

Sol :

Given : $x+\frac{1}{x}=4$

(i) Squaring on both sides

⇒$\left(x+\frac{1}{x}\right)^2=4^2$

⇒$x^2+2.x.\frac{1}{x}+\frac{1}{x^2}=16$

⇒$x^2+2+\frac{1}{x^2}=16$

⇒$x^2+\frac{1}{x^2}=16-2$

⇒$x^2+\frac{1}{x^2}=14$

(ii) $x^4+\frac{1}{x^4}$

∵We know that $x^2+\frac{1}{x^2}=14$

∴Squaring on both sides

⇒$\left(x^2+\frac{1}{x^2}\right)=14^2$

⇒$(x^2)^2+2.x^2..\frac{1}{x^2}+\left(\frac{1}{x^2}\right)^2=196$

⇒$x^4+2+\frac{1}{x^4}=196$

⇒$x^4+\frac{1}{x^4}=196-2$

⇒$x^4+\frac{1}{x^4}=194$

(iii) $x^3+\frac{1}{x^3}$

∵We know that $x+\frac{1}{x}=4$

Cubing on both sides

⇒$\left(x+\frac{1}{x}\right)^3=4^3$

⇒$x^3+3\times \frac{1}{x} \left(x+\frac{1}{x}\right)+\frac{1}{x^3}=64$

⇒$x^3+3\left(x+\frac{1}{x}\right)+\frac{1}{x^3}=64$

⇒$x^3+\frac{1}{x^3}=64-12$

=52

(iv) $x-\frac{1}{x}$

$\left(x-\frac{1}{x}\right)^2=x^2-2.x.\frac{1}{x}+\frac{1}{x^2}$

$=x^2+\frac{1}{x^2}-2$ ∵From (i) $x^2+\frac{1}{x^2}=14$

=14-2

$\left(x-\frac{1}{x}\right)^2=12$

$x-\frac{1}{x}=\sqrt{12}$

=√4×3

=2√3

Question 16

Sol :

Given : $x-\frac{1}{x}=5$

Squaring on both sides

⇒$\left(x-\frac{1}{x}\right)^2=5^2$

⇒$x^2-2.x.\frac{1}{x}+\frac{1}{x^2}=25$

⇒$x^2+\frac{1}{x^2}=25+2$

⇒$x^2+\frac{1}{x^2}=23$

Squaring on both sides

⇒$\left(x^2+\frac{1}{x^2}\right)^2=23^2$

⇒$(x^2)^2+2.x^2.\frac{1}{x^2}+\left(\frac{1}{x^2}\right)^2=529$

⇒$x^4+2+\frac{1}{x^4}=529$

⇒$x^4+\frac{1}{x^4}=529-2$

⇒$x^4+\frac{1}{x^4}=527$

Question 17

Sol :

Given : $x-\frac{1}{x}=\sqrt{5}$

(i) Squaring on both sides

⇒$\left(x-\frac{1}{x}\right)^2=(\sqrt{5})^2$

⇒$x^2-2.x.\frac{1}{x}+\frac{1}{x^2}=5$

⇒$x^2-2+\frac{1}{x^2}=5$

⇒$x^2+\frac{1}{x^2}=5+2$

⇒$x^2+\frac{1}{x^2}=7$

(ii) $x+\frac{1}{x}$

$\left(x+\frac{1}{x}\right)^2=x^3+2.x.\frac{1}{x}+\frac{1}{x^2}$

$=x^2+\frac{1}{x^2+2}$ [∵From (i)]

=7+2=9

⇒$\left(x+\frac{1}{x}\right)^2=9$

⇒$x+\frac{1}{x}=\sqrt{9}$

⇒$x+\frac{1}{x}=3$

(iii) $x^3+\frac{1}{x^3}$

⇒$\left(x+\frac{1}{x}\right)\left(x^2-x.\frac{1}{x}+\frac{1}{x^2}\right)$

⇒$\left(x+\frac{1}{x}\right)\left(x^2+\frac{1}{x^2}-1\right)$

⇒3(7-1)

⇒3×6=18

Question 18

Sol :

Given : $x+\frac{1}{x}=6$

Squaring on both sides

⇒$\left(x+\frac{1}{x}\right)^2=6^2$

⇒$x^2+2.x.\frac{1}{x}+\frac{1}{x^2}=36$

⇒$x^2+2+\frac{1}{x^2}=36$

⇒$x^2+\frac{1}{x^2}=36-2$

⇒34

(i) $x-\frac{1}{x}$

⇒$\left(x+\frac{1}{x}\right)^2=x^2-2.x\frac{1}{x}+\frac{1}{x^2}$

⇒$x^2+\frac{1}{x^2}-2$

⇒34-2

⇒$\left(x-\frac{1}{x}\right)^2=32$

⇒$x-\frac{1}{x}=\sqrt{32}$

⇒$x-\frac{1}{x}=\sqrt{16\times 2}$

⇒$x-\frac{1}{x}=4\sqrt{2}$

(ii) $x^2-\frac{1}{x^2}$

⇒$x^2-\frac{1}{x^2}=\left(x+\frac{1}{x}\right)\left(x-\frac{1}{x}\right)$

⇒$=6.4\sqrt{2}$

⇒24√2

Question 19

Sol :

Given : $x+\frac{1}{x}=2$

Squaring on both sides

⇒$\left(x+\frac{1}{x}\right)^2=2^2$

⇒$x^2+2.x.\frac{1}{x}+\frac{1}{x^2}=4$

⇒$x^2+2+\frac{1}{x^2}=4$

⇒$x^2+\frac{1}{x^2}=4-2$

⇒$x^2+\frac{1}{x^2}=2$...(i)

∴$x^3+\frac{1}{x^3}=\left(x+\frac{1}{x}\right)\left(x^2-x.\frac{1}{x}+\frac{1}{x^2}\right)$

$=2.\left(x^2+\frac{1}{x^2-1}\right)$

=2(2-1)

=2(1)

=2....(ii)

From (1) ⇒$x^2+\frac{1}{x^2}=2$

Squaring on both sides

⇒$\left(x^2+\frac{1}{x^2}\right)^2=2^2$

⇒$(x^2)^2+2.x^2.\frac{1}{x^2}+\left(\frac{1}{x^2}\right)^2=4$

⇒$x^4+2+\frac{1}{x^4}=4$

⇒$x^4+\frac{1}{x^4}=4-2$

=2..(iii)

From equation (i) , (ii) and (iii)

⇒$x^2+\frac{1}{x^2}=2$

⇒$x^3+\frac{21}{x^3}=2$

⇒$x^4+\frac{1}{x^4}=2$

∴$x^2+\frac{1}{x^2}=x^3+\frac{1}{x^3}=x^4+\frac{1}{x^4}$

Hence proved

Question 20

Sol :

Given : $x-\frac{2}{x}=3$

Cubing on both sides

⇒$\left(x-\frac{2}{x}\right)^3=3^3$

⇒$x^3-3.x.\frac{2}{x}\left(x-\frac{2}{x}\right)-\left(\frac{2}{x}\right)^3=27$

⇒$x^3-6\left(x-\frac{2}{x}\right)-\frac{8}{x^3}=27$

⇒$x^3-\frac{8}{x^3}-6(3)=27$

⇒$x^3-\frac{8}{x^3}=27+18$

=45

Question 21

Sol :

Given : a+2b=5

Cubing on both sides

⇒(a+2b)³=5³

⇒a³+3.a.2b(a+2b)+(2b)³=125

⇒a³+6ab(5)+8b³=125

⇒a³+30ab+8b³=125

⇒a³+8b³+30ab=125

Question 22

Sol :

Given : $a+\frac{1}{a}=P$

Cubing on both sides

⇒$\left(a+\frac{1}{a}\right)^3=p^3$

⇒$a^3+3.x.\frac{1}{x}\left(a+\frac{1}{a}\right)+\frac{1}{a^3}=p^3$

⇒$a^3+3(p)+\frac{1}{a^3}=p^3$

⇒$a^3+\frac{1}{a^3}=p^3-3p$

⇒$a^3+\frac{1}{a^3}=p\left(p^2-3\right)$

Question 23

Sol :

Given : $x^2+\frac{1}{x^2}=27$

⇒∵$\left(x-\frac{1}{x}\right)^2=x^2-2.x.\frac{1}{x}+\frac{1}{x^2}$

⇒$=x^2-2+\frac{1}{x^2}$

⇒$=x^2+\frac{1}{x^2}-2$

⇒=27-2

⇒=25

⇒$\left(x-\frac{1}{x}\right)^2=25$

⇒$x-\frac{1}{x}=\sqrt{25}$

⇒$x-\frac{1}{x}=5$

Question 24

Sol :

Given : $x^2+\frac{1}{x^2}=27$

Take $\left(x-\frac{1}{x}\right)^2=x^2-2.x.\frac{1}{x}+\frac{1}{x^2}$

⇒$=x^2-2+\frac{1}{x^2}$

⇒$=x^2+\frac{1}{x^2}-2$

⇒=27-2

⇒$\left(x-\frac{1}{x}\right)^2=25$

⇒$x-\frac{1}{x}=\sqrt{25}$

⇒$x-\frac{1}{x}=5$

∴$3x^3+5x-\frac{3}{x^3}-\frac{5}{x}$

⇒$3x^3-\frac{3}{x^3}+5x-\frac{5}{x}$

⇒$3\left(x^3-\frac{1}{x^3}\right)+5\left(x-\frac{1}{x}\right)$

⇒$3\left(x-\frac{1}{x}\right)\left(x^2+x.\frac{1}{x}+\frac{1}{x^2}\right)+5\left(x-\frac{1}{x}\right)$

⇒$3\left(x-\frac{1}{x}\right)\left(x^2+\frac{1}{x^2}+1\right)+5\left(x-\frac{1}{x}\right)$

⇒3(5)+(27+1)+5(5)

⇒15(28)+25

⇒420+25

⇒445

Question 25

Sol :

Given : $x^2+\frac{1}{25x^2}=8\frac{3}{5}$

$x^2+\left(\frac{1}{5x}\right)^2=\frac{4^3}{5}$

∴Let us consider

⇒$\left(x+\frac{1}{5x}\right)^2=x^2+2.x.\frac{1}{5x}+\left(\frac{1}{5x}\right)^2$

⇒$=x^2+\frac{2}{5}+\frac{1}{25x^2}$

⇒$=x^2+\frac{1}{25x^2}+\frac{2}{5}$

⇒$=\frac{43}{5}+\frac{2}{5}$

⇒$\left(x+\frac{1}{5x}\right)^2=\frac{43+2}{5}$

⇒$\left(x+\frac{1}{5x}\right)^2=\frac{47}{5}$

⇒$x+\frac{1}{5x}=\sqrt{\frac{47}{5}}$

Question 26

Sol :

Given : $x^2+\frac{1}{4x^2}=8$

$x^2+\left(\frac{1}{2x}\right)^2=8$

Let us consider

⇒$\left(x+\frac{1}{2x}\right)^2=x^2+2.x.\frac{1}{2x}+\left(\frac{1}{2x}\right)^2$

⇒$=x^2+1+\frac{1}{4x^2}$

⇒$=x^2+\frac{1}{4x^2}+1$

⇒=8+1

⇒$\left(x+\frac{1}{2x}\right)^2=9$

⇒$x+\frac{1}{2x}=\sqrt{9}$

⇒$x+\frac{1}{2x}=3$

∵$x^3+\left(\frac{1}{2x}\right)^3$

⇒$x^3+\frac{1}{3x^3}=\left(x+\frac{1}{2x}\right)\left(x^2-x.\frac{1}{2x}+\left(\frac{1}{2x}\right)^2\right)$

⇒$x^3+\frac{1}{8x^3}=\left(x+\frac{1}{2x}\right)\left(x^2+\frac{1}{4x^2}-1\right)$

=3(8-1)

=3(7)=21

Question 27

Sol :

Given : a²-3a+1=0

Dividing each term by a , we get

⇒$\frac{a^2}{a}-\frac{3a}{a}+\frac{1}{a}=0$

⇒$a-3+\frac{1}{a}=0$

⇒$a+\frac{1}{a}=3$

Now 

(i) $\left(a+\frac{1}{a}\right)^2=a^2+2.a.\frac{1}{a}+\frac{1}{a^2}$

⇒$\left(a+\frac{1}{a}\right)^2=a^2+2+\frac{1}{a^2}$

⇒$a^2+\frac{1}{a^2}=\left(a+\frac{1}{a}\right)^2-2$

=3²-2

=9-2=7

(ii) $a^3+\frac{1}{a^2}$

⇒$\left(a+\frac{1}{a}\right)\left(a^2-a.\frac{1}{a}+\frac{1}{a^2}\right)$

⇒$\left(a+\frac{1}{a}\right)\left(a^2+\frac{1}{a^2}-1\right)$

⇒(3)(7-1)

⇒3×6

⇒18

Question 28

Sol :

Given : $a=\frac{1}{a-5}$

⇒a(a-5)=1

⇒a²-5a=1

⇒a²-5a-1=0

(i) Dividing each term by a , we get

⇒$\frac{a^2}{a}-\frac{5a}{a}-\frac{1}{a}=0$

⇒$a-5-\frac{1}{a}=0$

∴$a-\frac{1}{a}=5$

(ii) Now $\left(a+\frac{1}{a}\right)$

∵$a-\frac{1}{a}=5$

Squaring on both sides

⇒$\left(a-\frac{1}{a}\right)^2=5^2$

⇒$a^2-2.a.\frac{1}{a}+\frac{1}{a^2}=25$

⇒$a^2+\frac{1}{a^2}=25-2$

⇒$a^2+\frac{1}{a^2}=23$

∴$\left(a+\frac{1}{a}\right)^2=a^2+2.a.\frac{1}{a}+\frac{1}{a^2}$

⇒$=a^2+\frac{1}{a^2}+2$

=23+2=25

∴$\left(a+\frac{1}{a}\right)^2=25$

$\left(a+\frac{1}{a}\right)=\sqrt{25}$

=5

(iii) $a^2-\frac{1}{a^2}=\left(a+\frac{1}{a}\right)\left(a-\frac{1}{a}\right)$

=5.5

=25

Question 29

Sol :

Given : $\left(x+\frac{1}{a}\right)^2=3$

⇒$x^2+2.x.\frac{1}{x}+\frac{1}{x^2}=3$

⇒$x^2+\frac{1}{x^2}=3-2$

⇒$x^2+\frac{1}{x^2}=1$

∵$x^3+\frac{1}{x^3}=\left(x+\frac{1}{x}\right)\left(x^2-x.\frac{1}{x}+\frac{1}{x^2}\right)$

$=\left(x+\frac{1}{x}\right)\left(x^2+\frac{1}{x^2}-1\right)$

$=\sqrt{3}(1-1)$

$=\sqrt{3}(0)=0$

Question 30

Sol :

Given : $x=5-2\sqrt{6}$

Squaring on both sides

$x^2=5$

Question 31

Sol :

Given : a+b+c=12

Squaring on both sides

⇒(a+b+c)²=12²

⇒a²+b²+c²+2(ab+bc+ca)=144

∵From given ab+bc+ca=22

∴a²+b²+c²+2(22)=144

⇒a²+b²+c²+44=144

⇒a²+b²+c²=144-44

⇒a²+b²+c²=100

Question 32

Sol :

Given : a+b+c=12

Squaring on both sides

⇒(a+b+c)²=12²

⇒a²+b²+c²+2(ab+bc+ca)=144

∵a²+b²+c²=100

⇒100+2(ab+bc+ca)=144

⇒2(ab+bc+ca)=144-100

⇒2(ab+bc+ca)=44

⇒$ab+bc+ca=\frac{44}{2}=22$

Question 33

Sol :

Given : a²+b²+c²=125

∴ab+bc+ca=50

∵(a+b+c)²=a²+b²+c²+2(ab+bc+ca)

=125+2(50)

=125+100

=225

∴(a+b+c)²=225

⇒a+b+c=√225

⇒a+b+c=15

Question 34

Sol :

Given : a+b-c=5

⇒a²+b²+c²=29

⇒(a+b+c)²=a²+b²+c²+2(ab-bc-ca)

⇒5²=29+2(ab-bc-ca)

⇒25=29+2(ab-bc-ca)

⇒25-29=2(ab-bc-ca)

⇒-4=2(ab-bc-ca)

⇒$ab-bc-ca=\frac{-4}{2}=-2$

Question 35

Sol :

Given : a-b=7

⇒a²+b²=85

⇒(a-b)²=a²+b²-2ab

⇒7²=85-2ab

⇒49=85-2ab

⇒2ab=85-49

⇒2ab=36

⇒$ab=\frac{36}{2}$

⇒ab=18

∴a³-b³=(a-b)(a²+ab+b²)

=7(a²+b²+ab)

=7(85+18)

=7(103)

=721

Question 36

Sol :

Given , the number is x

∴ x=y-3

∴x-y=-3

⇒y-x=3

and x²+y²=29

∴y-x=3

Squaring on both sides

⇒(y-x)²=3²

⇒y²+x²-2xy=9

⇒29-2xy=9

⇒2xy=20

⇒$xy=\frac{20}{2}$

⇒xy=10

Question 37

Sol :

Given : Sum of two numbers=8

Product of two numbers=15

Let, numbers be x and y

∴x+y=8

⇒xy=15

⇒x+y=8

∴x+y=8

Squaring on both sides

⇒(x+y)²=8²

⇒x²+y²+2xy=64

⇒x²+y²+2(15)=64

⇒x²+y²=64-30

⇒x²+y²=34

∴x³+y³

⇒(x+y)(x²-xy+y²)

⇒8(x²+y²-xy)

⇒8(34-8)

⇒8(26)

⇒208