ML Aggarwal Solution Class 9 Chapter 3 Expansions Exercise 3.1

 Exercise 3.1

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Q1 | Ex-3.1 | Class 9 | Expansions | ML Aggarwal | Chapter 3 | myhelper

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

Sol :

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

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

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Q2 | Ex-3.1 | Class 9 | Expansions | ML Aggarwal | Chapter 3 | myhelper

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Question 2

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

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

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

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

(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

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

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

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Q3 | Ex-3.1 | Class 9 | Expansions | ML Aggarwal | Chapter 3 | myhelper

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Question 3

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

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

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

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

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

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

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

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

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

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

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

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Q4 | Ex-3.1 | Class 9 | Expansions | ML Aggarwal | Chapter 3 | myhelper

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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

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Q5 | Ex-3.1 | Class 9 | Expansions | ML Aggarwal | Chapter 3 | myhelper

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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

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Q6 | Ex-3.1 | Class 9 | Expansions | ML Aggarwal | Chapter 3 | myhelper

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Question 6

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

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

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

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

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

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

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

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

Sol :

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

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

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

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

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

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

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Q7 | Ex-3.1 | Class 9 | Expansions | ML Aggarwal | Chapter 3 | myhelper

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Question 7

(i) (x+2)³

Sol :

Here , a=x, b=2

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

⇒x³+3(x²)(2)+3(x)(2²)+2³

⇒x³+6(x²)+3(x)(4)+8

⇒x³+6x²+12x+8

(ii) (2a+b)³

Sol :

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

⇒(2a)³+3.(2a)².b+3.2a.b²+b³

⇒8a³+3.4a².b+6ab²+b³

⇒8a³+12a²b+6ab²+b³

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Q8 | Ex-3.1 | Class 9 | Expansions | ML Aggarwal | Chapter 3 | myhelper

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Question 8

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

Sol :

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

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

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

⇒$27x^3+3.(9x^2).\dfrac{1}{x}+9.(x).\dfrac{1}{x^2}+\dfrac{1}{x^3}$

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

(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

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Q9 | Ex-3.1 | Class 9 | Expansions | ML Aggarwal | Chapter 3 | myhelper

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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+135xy²-27y³

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

Sol :

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

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

⇒$8x^3-3.(4x^2).\dfrac{1}{3y}+3.(2x).\dfrac{1}{9y^2}-\dfrac{1}{27y^3}$

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

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Q10 | Ex-3.1 | Class 9 | Expansions | ML Aggarwal | Chapter 3 | myhelper

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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

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Q11 | Ex-3.1 | Class 9 | Expansions | ML Aggarwal | Chapter 3 | myhelper

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Question 11

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

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

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

⇒$2a^2+\dfrac{2}{a^2}$

⇒$2\left(a^2+\dfrac{1}{a^2}\right)$

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

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

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

⇒2+2

⇒4

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Q12 | Ex-3.1 | Class 9 | Expansions | ML Aggarwal | Chapter 3 | myhelper

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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

(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)

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Q13 | Ex-3.1 | Class 9 | Expansions | ML Aggarwal | Chapter 3 | myhelper

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Question 13

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

⇒7p(7p-9q)+9q(7p-9q)

⇒49p²-63pq+63pq-81q²

⇒49p²-81q²

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

⇒$(2x)^2-\left(\dfrac{3}{x}\right)^2$

⇒$(2x)^2-\left(\dfrac{3}{x}\right)^2$

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

∴ $4x^2-\dfrac{9}{x^2}$

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Q14 | Ex-3.1 | Class 9 | Expansions | ML Aggarwal | Chapter 3 | myhelper

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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

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Q15 | Ex-3.1 | Class 9 | Expansions | ML Aggarwal | Chapter 3 | myhelper

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Question 15

(i) $\left(x+\dfrac{2}{x}-3\right)\left(x-\dfrac{2}{x}-3\right)$

⇒$\left((x-3)+\dfrac{2}{x}\right)\left((x-3)-\dfrac{2}{x}\right)$

It is in the form of (a+b)(a-b)=a²-b²

∵a=x-3 ; $b=\dfrac{2}{x}$

⇒$(x-3)^2-\left(\dfrac{2}{x}\right)^2$

⇒$x^2-2.(x).3+3^2-\dfrac{4}{x^2}$

⇒$x^2-6x+9-\dfrac{4}{x^2}$

(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

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Q16 | Ex-3.1 | Class 9 | Expansions | ML Aggarwal | Chapter 3 | myhelper

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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

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Q17 | Ex-3.1 | Class 9 | Expansions | ML Aggarwal | Chapter 3 | myhelper

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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)=a³+b³

∴Here, a=2p ; b=3q

∴(2p)³+(3q)³

⇒8p³+27q³

(ii) $\left(x+\dfrac{1}{x}\right)\left(x^2-1+\dfrac{1}{x^2}\right)$

Sol :

It is in the form of (a+b)(a²-ab+b²)=a³+b³

∴Here a=x ; $b=\dfrac{1}{x}$

∴$x^3+\dfrac{1}{x^3}$

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Q18 | Ex-3.1 | Class 9 | Expansions | ML Aggarwal | Chapter 3 | myhelper

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Question 18

(i) (3p-4q)(9p²+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) $\left(x-\dfrac{3}{x}\right)\left(x^2+3+\dfrac{9}{x^2}\right)$

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

It is in the form of (a-b)(a²+ab+b²) is a³-b³

∴Here, a=x ; $b=\dfrac{3}{x}$

⇒$x^3-\dfrac{27}{x^3}$

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Q19 | Ex-3.1 | Class 9 | Expansions | ML Aggarwal | Chapter 3 | myhelper

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Question 19

$(2x + 3y + 4z) (4x^2 + 9y^2 + 16z^2 – 6xy – 12yz – 8zx)$

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

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Q20 | Ex-3.1 | Class 9 | Expansions | ML Aggarwal | Chapter 3 | myhelper

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Question 20

Find the product of the following:

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

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

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Q21 | Ex-3.1 | Class 9 | Expansions | ML Aggarwal | Chapter 3 | myhelper

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Question 21

Find the coefficient of $x^2$ and x in the product of (x – 3) (x + 7) (x – 4)

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

∴ Coefficient of $x^2$ is 0, coefficient of x is -37

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Q22 | Ex-3.1 | Class 9 | Expansions | ML Aggarwal | Chapter 3 | myhelper

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Question 22

If $a^2 + 4a + x = (a + 2)^2$, find the value of x.

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

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Q23 | Ex-3.1 | Class 9 | Expansions | ML Aggarwal | Chapter 3 | myhelper

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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

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Q24 | Ex-3.1 | Class 9 | Expansions | ML Aggarwal | Chapter 3 | myhelper

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Question 24

(i) 99²

⇒(100-1)²

⇒(100)²-2.100.1+1²

⇒10000-200+1

⇒9801

(ii) (9997)²

⇒(10000-3)²

⇒10000²-2.10000.9+3²

⇒100000000-60000+9

⇒99940009

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

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Q25 | Ex-3.1 | Class 9 | Expansions | ML Aggarwal | Chapter 3 | myhelper

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Question 25

(i) (103)³

⇒(100+3)³

∴It is in the form of 

(a+b)³=a³+3a²b+3ab²+b³

∴Here a=100 ; b=3

⇒(100)³+3×(100)²×3+3×100×(3)²+3³

⇒1000000+90000+2700+27

⇒1092727

(ii) 99³

⇒(100-1)³

∴It is in the form of 

(a-b)³=a³-3a²b+3ab²-b³

∴Here a=100 ; b=1

⇒(100)³-3×(100)²×1+3×100×(1)²-(1)³

⇒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+0.3+0.001

⇒1030.301

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Q26 | Ex-3.1 | Class 9 | Expansions | ML Aggarwal | Chapter 3 | myhelper

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Question 26

If 2a – b + c = 0, prove that $4a^2 – b^2 + c^2 + 4ac = 0$

Hint: 2a – b + c = 0 ⇒ $(2a + c)^2= b^2$

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

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Q27 | Ex-3.1 | Class 9 | Expansions | ML Aggarwal | Chapter 3 | myhelper

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Question 27

If a + b + 2c = 0, Prove that $a^3 + b^3 + 8c^3 = 6abc$

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

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Q28 | Ex-3.1 | Class 9 | Expansions | ML Aggarwal | Chapter 3 | myhelper

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Question 28

If a + b + c = 0, then find the value of  $\dfrac{a^2}{bc} + \dfrac{b^2}{ca} + \dfrac{c^2}{ab}$

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

⇒$\dfrac{a^3+b^3+c^3}{abc}=3$
⇒$\dfrac{a^3}{abc}+\dfrac{b^3}{abc}+\dfrac{c^3}{abc}=3$

∴$\dfrac{a^2}{bc}+\dfrac{b^2}{ac}+\dfrac{c^2}{ab}=3$

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Q29 | Ex-3.1 | Class 9 | Expansions | ML Aggarwal | Chapter 3 | myhelper

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Question 29

If x + y = 4, then find the value of $x^3 + y^3 + 12xy – 64$.

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

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Q30 | Ex-3.1 | Class 9 | Expansions | ML Aggarwal | Chapter 3 | myhelper

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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

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Q31 | Ex-3.1 | Class 9 | Expansions | ML Aggarwal | Chapter 3 | myhelper

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Question 31

Using suitable identity, find the value of:

$\dfrac{86×86×86+14×14×14}{86×86-86×14+14×14}$

Sol :

Given $\dfrac{86\times 86\times 86+14\times 14\times 14}{86\times 86-86\times 14 +14\times 14}$

∴It is in the form of $\dfrac{a^3+b^3}{a^2-ab+b^2}=(a+b)$

∴$\dfrac{(86)^3+(14)^3}{86^2-86\times 14+14^2}=86+14$

=100