ML Aggarwal Solution Class 9 Chapter 3 Expansions Exercise 3.1

 Exercise 3.1

Question 1

(i) (2x+7y)2

It is in the form of (a+b)2=a2+2ab+b2

∴a=2x, b=7y

∴(2x+7y)2=(2x)2+2.2x.7y+(7y)2

=4x2+28xy+49y2


(ii) $\left(\frac{1}{2}x+\frac{2}{3}y\right)^2$

Sol :

$\left(\frac{1}{2}x \right)^2+2\times \frac{1}{2} \times x \times \frac{2}{3}\times y+\left(\frac{2}{3} y\right)^2$

$\frac{x^2}{4}+\frac{2xy}{3}+\frac{4}{9}y^2$


Question 2

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

It is in the form of $(a+b)^2=a^2+2ab+b^2$

∴$(3x)^2+2.3x.\frac{1}{2x}+\left(\frac{1}{2x}\right)^2$

$9x^3+3+\frac{1}{4x^2}$


(ii) $(3x^2y+5z)^2$

It is in the form of $(a+b)^2=a^2+2ab+b^2$

Here $a=3x^2y$ , b=5x

$(3x^2y)^2+2.3x^2y.5z+(5z)^2$

$9x^4y^2+30x^2yz+25z^2$


Question 3

(i) $\left(3x-\frac{1}{2x}\right)^2$

It is in the form of $(a-b)^2=a^2-2ab+b^2$

Here, a=3x ; $b=\frac{1}{2x}$

$(3x)^2-2.3x.\frac{1}{2x}+\left(\frac{1}{2x}\right)^2$

$3^2-x^2-3+\frac{1}{2^2x^2}$

$9x^2-3+\frac{1}{4x^2}$


(ii) $\left(\frac{1}{2}x-\frac{3}{2}y\right)^2$

It is in the form of $(a-b)^2=a^2-2ab+b^2$

Here, $a=\frac{1}{2}x$ ; $b=\frac{3}{2}y$

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

$\frac{x^2}{4}-\frac{3xy}{2}+\frac{9y^2}{4}$


Question 4

(i) (x+3)(x+5)

⇒x(x+5)+3(x+5)

⇒$x^2+5x+3x+15$

⇒$x^2+8x+15$


(ii) (x+3)(x-5)

⇒x(x-5)+3(x-5)

⇒x.x-x.5+3.x-3.5

⇒x2-5x+3x-15

⇒x2-2x-15


(iii) (x-7)(x+9)

⇒x(x+9)-7(x+9)

⇒x.x+9.x-7.x-7.9

⇒x2+9x-7x-63

⇒x2+2x-63


(iv) (x-2y)(x-3y)

⇒x(x-3y)-2y(x-3y)

⇒x.x-x.3y-2y.x+2y.3y

⇒x2-3xy-2xy+6y2

⇒x2-5xy+6y2


Question 5

(i) (x-2y-z)2

It is in the form of (a+b+c)2=a2+b2+c2+2(ab+bc+ca)

Here, a=x, b=-2y, c=-z

⇒x2+(-2y)2+(-z)2+2(x(-2y)+(-2y)(-z)+(-z)x)

⇒x2+4y2+z2+2(-2xy+2yz-zx)

⇒x2+4y2+z2+4yz-4xy-2zx


(ii) (2x-3y+4z)2

It is in the form of (a+b+c)2=a2+b2+c2+2(ab+bc+ca)

Here a=2x, b=-3y, c=4z

⇒(2x)2+(-3y)+(4z)+2(2x.(-3y)+(-3y)(4z)+(4z)(2x))

⇒4x2+4y2+16z2+2(-6xy-12yz+8xz)

⇒4x2+4y2+16z2-12xy-24yz+16xz


Question 6

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

It is in the form of (a+b+c)2=a2+b2+c2+2(ab+bc+ca)

Here, a=2x , $b=\frac{3}{x}$ , c=-1

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

⇒$4x^2+\frac{9}{x^2}+1+2\left(6-\frac{3}{x}-2x\right)$

⇒$4x^2+\frac{9}{x^2}+1+12-\frac{6}{x}-4x$

⇒$4x^2+\frac{9}{x^2}-\frac{6}{x}-4x+13$


(ii) $\left(\frac{2}{3}x-\frac{3}{2x}-1\right)^2$

Sol :

It is in the form of (a+b+c)2=a2+b2+c2+2(ab+bc+ca)

Here, $a=\frac{2}{3}x,b=\frac{-3}{2x}$, c=-1

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

⇒$\frac{4}{9}x^2-\frac{9}{4x^2}+1+2\left[-1+\frac{3}{2x}-\frac{2}{3}x\right]$

⇒$\frac{4}{9}x^2-\frac{9}{4x^2}+1+2+\frac{6}{2x}-\frac{4x}{3}$

⇒$\frac{4}{9}x^2-\frac{9}{4x^2}+\frac{3}{x}-\frac{4x}{3}-1$


Question 7

(i) (x+2)3

Sol :

It is in the form of (a+b+c)2=a2+b2+c2+2(ab+bc+ca)

Here , a=x, b=2

∴⇒x3+3.x2.2+3.x.2+23

⇒x3+6.x2+3x.4+8

⇒x3+6x2+12x+8


(ii) (20+b)3

Sol :

⇒(2a)3+3.(2a)2.b+3.2a.b2+b3

⇒8a3+3.4a2.b+6ab2+b3

⇒8a+12a2b+6ab2+b3


Question 8

(i) $\left(3x+\frac{1}{x}\right)^3$

Sol :

It is in the form of (a+b+c)3=a3+3a2b+3ab2+b3

a=3x ; $b=\frac{1}{x}$

∴$(3x)^3+3.(3x)^2.\frac{1}{x}+3.3x.\left(\frac{1}{x}\right)^2+\left(\frac{1}{x}\right)^3$

⇒$27x^3+3.9x^2\frac{1}{x}+9.x.\frac{1}{x^2}+\frac{1}{x^3}$

⇒$27x^3+27x+\frac{9}{x}+\frac{1}{x^3}$


(ii) (2x-1)3

Sol :

It is in the form of (a-b)3=a3-3a2b+3ab2-b3

Here, a=2x, b=1

∴(2x)3-3(2x)2.1+3(2x)(1)2-(1)3

⇒8x3-3.4x2+6x-1

⇒8x3-12x2+6x-1


Question 9

(i) (5x-3y)3

Sol :

It is in the form of (a-b)3=a3-3a2b+3ab2-b3

a=5x ; b=3y

∴ (5x)3-3.(5x)2.3y+3.5x.(3y)2-(3y)3

⇒125x3-3.25x2.3y+3.5x.9y2-27y3

⇒125x3-225x2y+135y2.x-27y3


(ii) $\left(2x-\frac{1}{3y}\right)^3$

Sol :

⇒$(2x)^3-3.(2x)^2.\frac{1}{3y}+3.2x.\left(\frac{1}{3y}\right)^2-\left(\frac{1}{3y}\right)^3$

⇒$8x^3-3.4x^2.\frac{1}{3y}+3.2x.\frac{1}{9y^2}-\frac{1}{27y^3}$

⇒$8x^3-\frac{4x^2}{y}+\frac{2x}{3y^2}-\frac{1}{27y^3}$


Question 10

(i) (a+b)2+(a-b)2

⇒a2+2ab+b2+a2-2ab+b2

⇒2a2+2b2

⇒2(a2+b2)


(ii) (a+b)2-(a-b)2

⇒(a2+2ab+b2)-(a2-2ab+b2)

⇒a2+2ab+b2-a2+2ab-b2

⇒2ab+2ab

⇒4ab


Question 11

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

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

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

⇒$2a^2+\frac{2}{a^2}$

⇒$2\left(a^2+\frac{1}{a^2}\right)$


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

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

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

⇒2+2

⇒4


Question 12

(i) (3x-1)2-(3x-2)(3x+1)

⇒(3x)2-2.3x.1+12-3x(3x+1)+2(3x+1)

⇒9x2-6x+1-9x2-3x+6x+2

⇒3x+3

⇒3(x+1)


(ii) (4x+3y)2-(4x-3y)2-48

⇒(4x)2+2.3y.4x+(3y)2-((4x)2-2.4x.3y+(3y)2)-48

⇒16x2+24xy+9y2-16x2+24xy-9y2-48

⇒48xy-48

⇒48(xy-1)


Question 13

(i) (7p+9q)(7p-9q)

⇒7p(7p-9q)+9q(7p-9q)

⇒49p2-63pq+63pq-81q2

⇒49p2-81q2


(ii) $\left(2x-\frac{3}{x}\right)\left(2x+\frac{3}{x}\right)$

⇒$(2x)^2-\left(\frac{3}{x}\right)^2$

⇒$(2x)^2-\left(\frac{3}{x}\right)^2$

⇒Since it is in the form of (a+b)(a-b)=a2-b2

∴ $4x^2-\frac{9}{x^2}$


Question 14

(i) (2x-y+3)(2x-y-3)

⇒((2x-y)+3)((2x-y)-3)

It is in the form of (a+b)(-a-b)=a2-b2

∴(2x-y)2-32
⇒(2x)2-2.2x.y+y2-9
⇒4x2-4xy+y2-9

(ii) (3x+y-5)(3x-y-5)
⇒(3x+(y-5))(3x-(y+5))
⇒[(3x-5)+y][(3x-5)-y]
⇒It is in the form of (a+b)(a-b)=a2-b2
∵a=3x-5 ; b=y
∴(3x-5)2-y2
⇒(3x)2-2.3x.5+52-y2
⇒9x2-30x+25-y2

Question 15

(i) $\left(x+\frac{2}{x}-3\right)\left(x-\frac{2}{x}-3\right)$
⇒$\left((x-3)+\frac{2}{x}\right)\left((x-3)-\frac{2}{x}\right)$
It is in the form of (a+b)(a-b)=a2-b2
∵a=x-3 ; $b=\frac{2}{x}$
⇒$(x-3)^2-\left(\frac{2}{x}\right)^2$
⇒$x^2-2.x.3+3^2-\frac{4}{x^2}$
⇒$x^2-6x+9-\frac{4}{x^2}$

(ii) (5-2x)(5+2x)(25+4x2)
It is in the form of (a+b)(a-b)=a2-b2
∴(52-(2x)2)(25+4x2)
⇒(25-4x2)(25+4x2)
⇒(25)2-(4x2)2
⇒625-16x4


Question 16

(i) (x+2y+3)(2y+x+7)
⇒x(2y+x+7)+2y(2x+x+7)+3(2y+x+7)
⇒2xy+x2+7x+4y2+2xy+14y+6y+3x+21
⇒x2+4y2+4xy+10x+20y+21

(ii) (2x+y+5)(2x+y-9)
⇒2x(2x+y-9)+y(2x+y-9)+5(2x+y-9)
⇒4x2+2xy-18x+2xy+y2-9y+10x+5y-45
⇒4x2+y2+4xy-8x-4y-45

(iii) (x-2y-5)(x-2y+3)
⇒x(x-2y+3)-2y(x-2y+3)-5(x-2y+3)
⇒x2-2xy+3x-2xy+4y2-6y-5x+10y-15
⇒x2+4y2-4xy-2x-4y-15

(iv) (3x-4y-2)(3x-4y-6)
⇒3x(3x-4y-6)-4y(3x-4y-6)-2(3x-4y-6)
⇒9x2-12xy-18x-12xy+16y2+24y-6x+8y+12
⇒9x2+16y2-24xy-24x+32y+12


Question 17

(i) (2p+3q)(4p2-6pq+9q2)
Sol :
(2p+3q)((2p)2-2p.3q+(3q)2)
It is in the form of (a+b)(a2-ab)b2) is
a3+b3
∴Here, a=2p ; b=3q
∴(2p)3+(3q)3
⇒8p3+27q3

(ii) $\left(x+\frac{1}{x}\right)\left(x^2-1+\frac{1}{x^2}\right)$
Sol :
It is in the form of (a+b)(a2-ab+b2) is a3+b3
∴Here a=x ; $b=\frac{1}{x}$
∴$x^3+\frac{1}{x^3}$

Question 18

(i) (3p-4q)(4p2+12pq+16q2)
(3p-4q)((3p)2+3p.4q+(4q)2)
It is in the form of (a-b)(a2+ab+b2) is a3-b3
∴Here 3p=a ; b=4q
∴(3p)3-(4q)3
⇒27p3-64q3

(ii) $\left(x-\frac{3}{x}\right)\left(x^2+3+\frac{9}{x^2}\right)$
∴$\left(x-\frac{3}{x}\right)\left(x^2+x.\frac{3}{x}+\left(\frac{3}{2}\right)^2\right)$
It is in the form of (a-b)(a2+ab+b2) is a3-b3
∴Here, a=x ; $b=\frac{3}{x}$
⇒$x^3-\frac{27}{x^3}$

Question 19

Sol :
Given (2x+3y+4z)(4x2+9y2+16z2-6xy-12yz-8zx)
⇒(2x+3y+4z)((2x)2+(3y)2+(4z)2-2x.3y-3y.4z-4z.2x)
∴It is in the form of (a+b+c)(a2+b2+c2-ab-bc-ca)=a3+b3+c3-3abc
∴Here a=2x , b=3y , c=4z
∴(2x)2+(3y)2+(4z)3-3.2x.3y.4z
⇒8x3+27y3+64z3-72xyz

Question 20

Sol :
(i) (x+1)(x+2)(x+3)
[x(x+2)+1(x+2)](x+3)
⇒(x2+2x+x+2)(x+3)
⇒(x2+3x+2)(x+3)
⇒(x2+3x+2)(x)+(x2+3x+2)3
⇒x3+5x2+2x+3x2+9x+6
⇒x3+6x2+11x+6


(ii) (x-2)(x-3)(x+4)
⇒[x(x-3)-2(x-3)](x+4)
⇒(x2-3x-2x+6)(x+4)
⇒(x2-5x+6)(x+4)
⇒(x2-5x+6)x+(x2-5x+6)4
⇒x3-5x2+6x+4x2-20x+24
⇒x3-x2-14x+24

Question 21

Sol :
⇒Given : (x-3)(x+7)(x-4)
⇒(x(x+7)-3(x+7))(x-4)
⇒(x2+7x-3x-21)(x-4)
⇒(x2+4x-21)x-4(x2+4x-21)
⇒x3+4x2-21x-4x2-16x+84
⇒x3-37x+84

Question 22

Sol :
Given : 
⇒a2+4a+x=(a+2)2
⇒a2+4a+x=a2+2.a.2+22
⇒a2+4a+x=a2+4a+4
⇒x=a2+4a+4-a2-4a
⇒x=4

Question 23

(i) (101)2
⇒(100+1)2
⇒(100)2+2.100.1+12
⇒10000+200+1
⇒10201

(ii) (1003)2
⇒(1000+3)2
⇒(1000)2+2.1000.3+32
⇒1000000+6000+9
⇒1006009

(iii) (10.2)2
⇒(10+0.2)2
⇒(10)2+2.10×0.2+(0.2)2
⇒100+4+0.04
⇒104.04


Question 24

(i) 992
⇒(100-1)2
⇒(100)2-2.100.1+12
⇒10000-200+1
⇒9801

(ii) (9997)2
⇒(1000-3)2
⇒10002-2.1000.9+32
⇒1000000-6000+9
⇒994009
In this we used the (a-b)2
formulae i.e. (a2+2ab+b2)

(iii) (9.8)2
⇒(10-0.2)2
⇒102-2×10×0.2+(0.2)2
⇒100-4+0.4
⇒96.04

Question 25

(i) (103)3
⇒(100+3)2
∴It is in the form of 
(a+b)3=a3+3a2b+3ab2+b3
∴Here a=100 ; b=3
⇒(100)3+3.(100)2.3+33
⇒1000000+90000+2700+27
⇒1092727

(ii) 993
⇒
⇒100
⇒1000000-30000+300-1
⇒970299

(iii) (10.1)3
⇒(10+0.1)3
⇒103+3.102.(0.1)+3.10.(0.1)2+(0.1)3
⇒1000+30+3+0.01
⇒1030.301

Question 26

Sol :
Given : 2ab+c=0
⇒(2a+c)=0
Squaring both sides
⇒(2a+c)2=b2
⇒(2a)2+2.2a.c+c2=b2
⇒4a2+4ac+c2=b2
⇒4a2-b2+c2+4ac=0
Hence proved

Question 27

Sol :
⇒Given : a+b+2c=0
⇒a+b=-2c...(i)
⇒Cubing on both sides
⇒(a+b)3=(-2c)3
⇒a3+b3+3a2b+3ab2=-8c3
⇒a3+b3+3ab(a+b)=-8c3 [from (i)]
⇒a3+b3+3ab(-2c)=-8c3
⇒a3+b3-6abc=-8c3
⇒a3+b3+8c3=6abc
Hence proved

Question 28

Sol :
⇒Given a+b+c=0
⇒a+b=-c...(i)
Cubing on both sides
⇒(a+b)3=(-c)3
⇒a3+b3+3a2b+3ab2=-c3
⇒a3+b3+3ab(a+b)=-c3
⇒a3+b3+3ab(-c)=-c3
⇒a3+b3-3abc=-c3
⇒a3+b3+c3=3abc
⇒$\frac{a^3+b^3+c^3}{abc}=3$
⇒$\frac{a^3}{abc}+\frac{b^3}{abc}+\frac{c^3}{abc}=3$
∴$\frac{a^2}{bc}+\frac{b^2}{ac}+\frac{c^2}{ab}=3$

Question 29

Sol :
⇒Given x+y=4
Cubing on both sides
⇒(x+y)3=43
⇒x3+3x2y+3xy2+y3=64
⇒x3+3xy(x+y)+x3+
⇒x3+3xy(4)+y3=64
⇒x3+12xy+y3=64
⇒x3+y3+12xy-64=0

Question 30

(i) (27)3+(-17)3+(-10)3
∴If a+b+c=0; then a3+b3+c3=3abc
∴Here a=27 , b=17 , c=-10
∴27-17-10=0
∴a3+b3+c3=3abc
=3.27(-17)(-10)
=13770

(ii) (-28)3+153+133
If a+b+c=0 ; then a3+b3+c3=3abc
⇒-28+15+13=0
∴⇒a3+b3+c3=3abc
=3(-28)(15)(13)
=-16380

Question 31

Sol :
Given $\frac{86\times 86\times 86+14\times 14\times 14}{86\times 86-86\times 14 +14\times 14}$
∴It is in the form of $\frac{a^3+b^3}{a^2-ab+b^2}=(a+b)$
∴$\frac{(86)^3+(14)^3}{86^2-86.14+14^2}=86+14$
=100

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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 =8 2 ⇒x 2 -2.xy+y 2 =64 ⇒x 2 -2(5)+y 2 =64 ⇒x 2 +y 2 =64+10 ⇒x 2 +y 2 =74 Question 2 Sol : ⇒Given : x+y=10...(i) ⇒xy=21....(ii) Squaring both sides in equation (i) ⇒(x+y) 2 =10 2 ⇒x 2 +2xy+y 2 =100 ⇒x 2 +2(21)+y 2 =100 ⇒x 2 +4 2 +y 2 =100 ⇒x 2 +y 2 =100-42 ⇒x 2 +y 2 =58 2(x 2 +y 2 )=2×58=116 Question 3 Sol : ⇒Given : 2a+3b=7..(i) ⇒ab=2....(ii) Squaring both sides in equal (i) ⇒(2a+3b) 2 =7 2 ⇒(2a) 2 +2.2a+3b+(3b) 2 =49 ⇒4a 2 +12ab+9b 2 =49 ⇒4a 2 +12(2)+9b 2 =49 ⇒4a 2 +24+9b 2 =49 ⇒4a 2 +9b 2 =49-24 ⇒4a 2 +9b 2 =25 Question 4 Sol : ⇒Given : 3x-4y=16...(i) ⇒xy=4..(ii) Squaring on both sides in equation (i) ⇒(3x-4y) 2 =16 2 ⇒(3x) 2 -2.3x.4y+(4y) 2 =256 ⇒9x 2 -24xy+16y 2 =256 ⇒9x 2 -24(4)+16y 2 =256 ⇒9x 2 -96+16y 2 =256 ⇒9x 2 +16y 2 =256+96 ⇒9x 2 +16y 2 =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) ⇒S...
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