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# KC Sinha Mathematics Solution Class 9 Chapter 4 Algebraic identities exercise 4.5

Page 4.34

### Exercise 4.5

Type 1

Problems based on factorization of algebraic expressions, expressible as the sum or difference of two cubes.

WORKING RULE:

1. if in the given expression, any factor is common in each term , then take out the common factors.

Now , use the following identities whichever is required

x3+y3=(x+y)(x2-xy+y2)

x3-y3=(x-y)(x2+xy+y2)

2. Replace y be in x3+y3 to get x3-y3

#### Question 1

a3+27b3

Sol:

⇒(a)3+(3b)3

Using identity:

x3+y3=(x+y)(x2-xy+y2)

⇒(a+3b)[a2-(a)(3b)+(3b)2]

⇒(a+3b)(a2-3ab+9b2)

#### Question 2

x3+125

Sol:

⇒x3+53

Using identity:

x3+y3=(x+y)(x2-xy+y2)

⇒(x+5)[x2-(x)(5)+52]

⇒(x+5)(x2-5x+25)

#### Question 3

8a3+27b3

Sol:

⇒(2a)3+(3b)3

Using identity:

x3+y3=(x+y)(x2-xy+y2)

⇒(2a+3b)[(2a)2-(2a)(3b)+(3b)2]

⇒(2a+3b)(4a2+9b2-6ab)

#### Question 4

x5+27x2

[Hint: x5+27x2=x2(x3+27)]

Sol:

[Taking common x2]

⇒x2(x3+27)

⇒x2(x3+33)

Using identity:

x3+y3=(x+y)(x2-xy+y2)

⇒x2(x+3)(x2-3x+32)

⇒x2(x+3)(x2-3x+9)

#### Question 5

ab7+ba7

[Hint: ab7+ba7=ab(a6+b6)=ab[(a2)3+(b2)3]]

Sol:

[Taking common ab]

⇒ab(b6+a6)

⇒ab[(b2)3+(a2)3]

⇒ab[(a2)3+(b2)3]

Using identity:

x3+y3=(x+y)(x2-xy+y2)

⇒ab[(a2+b2)][(a2)2-(a2)(b2)+(b2)2]

⇒ab(a2+b2)(a4-a2b2+b4)

#### Question 6

8(a+b)3 + 27(b+c)3

Sol:

[Hint: Given expression {2(a+b)}]3+{3(b+c)}3

={2(a+b)+3(b+c)}{4(a+b)2-6(a+b)(b+c)+9(b+c)2}

=(2a+5b+3c){(a+b)(4a+4b-6b-6c)+9(b2+c2+2bc)}

=(2a+5b+3c){(a+b)(4a-2b-6c)+9b2+9c2+18bc}

=(2a+5b+3c)(4a2+7b2+9c2+2ab+12bc-6ac)

#### Question 7

Sol:

Using identity:

x3+y3=(x+y)(x2-xy+y2)

#### Question 8

a3+b3+a+b

Sol:

⇒a3+b3+a+b

Using identity:

x3+y3=(x+y)(x2-xy+y2)

⇒(a+b)(a2-ab+b2)+(a+b)

[Taking common (a+b)]

⇒(a+b)[(a2-ab+b2)+1]

⇒(a+b)(a2-ab+b2+1)

#### Question 9

x3+y3+2x2-2y2

Sol:

⇒(x3+y3)+2(x2-y2)

Using identity:

x3+y3=(x+y)(x2-xy+y2)

⇒[(x+y)(x2-xy+y2)]+2(x2-y2)

Using identity:

a2-b2=(a+b)(a-b)

⇒(x+y)(x2-xy+y2)+2(x+y)(x-y)

[Taking common (x+y)]

⇒(x+y)[(x2-xy+y2)+2(x-y)]

⇒(x+y)[x2-xy+y2+2x-2y]

⇒(x+y)(x2+y2-xy+2x-2y)

#### Question 10

64x3-27y3

Sol:

⇒64x3-27y3

⇒(4x)3-(3y)3

Using identity:

x3-y3=(x-y)(x2+xy+y2)

⇒(4x-3y)[(4x)2-(4x)(3y)+(3y)2]

⇒(4x-3y)(16x2+12xy+9y2)

#### Question 11

(a+b)3-(2)3

[Hint: Given expression=(a+b-2)[(a+b)2+2(a+b)+4]]

Sol:

Using identity:

x3-y3=(x-y)(x2+xy+y2)

⇒(a+b-2)[(a+b)2+2(a+b)+22]

⇒(a+b-2)[(a+b)(a+b+2)+4]

#### Question 12

x3y3-512

[Hint: Given expression =(xy)3-(8)3 ]

Sol :

⇒(xy)3-83

Using identity:

x3-y3=(x-y)(x2+xy+y2)

⇒(xy-8)[(xy)2+(8)(xy)+(8)2]

⇒(xy-8)(x2y2+8xy+64)

#### Question 13

Sol :

Expression

Using identity:

x3-y3=(x-y)(x2+xy+y2)

Page 4.35

Type 2

Problems based on the identity:

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

Category A: Problems based on factorization of the polynomials of the form a3+b3+c, when a+b+c=0

WORKING RULE:

Find the algebric sum of the terms. If it is zero, then required factors will be 3×product of three terms.

Thus if a+b+c=0 , then a3+b3+c3=3abc

Category B: Problems based on factorization of the polynomials of the form a3+b3+c-3abc

WORKING RULE:

1. If polynomial is of the form a3+b3+c-3abc , then use identity a3+b3+c3-3abc=(a+b+c)(a2+b2+c2-ab-bc-ca)

2. If polynomial can be converted in the form a3+b3+c-3abc , then convert the given expression in this form and then factorize.

3. Use the following formulae whichever is required :

(i) a2+b2+c2-ab-bc-ca=1/2{(a-b)2+(b-c)2+(c-a)2}

(ii) (a+b+c)3=a3+b3+c3+3(a+b)(b+c)(c+a)

(iii) a3+b3+c3=(a+b+c)3-3(a+b)(b+c)(c+a)

(iv) (a+b+c)3-a3-b3-c3=3(a+b)(b+c)(c+a)

#### Question 14

Fill up the blanks.

(i) If x+y+z=0 , factors of x3+y3+z3 will be __

Sol :

Using identity :

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

[Given x+y+z=0]

⇒x3+y3+z3-3xyz=(0)(x2+y2+z2-xy-yz-zx)

⇒x3+y3+z3-3xyz=0

⇒x3+y3+z3=3xyz

∴ Factors of x3+y3+z3 is

⇒3xyz

(ii) Factors of (a-b)3+(b-c)3+(c-a)3 will be __

Sol :

⇒(a-b)3+(b-c)3+(c-a)3

Using identity :

⇒a3+b3+c3=(a+b+c)3-3(a+b)(b+c)(c+a)

where a=(a-b) , b=(b-c) , c=(c-a)

⇒[(a-b)+(b-c)+(c-a)]-3[(a-b)+(b-c)][(b-c)+(c-a)][(c-a)+(a-b)]

⇒[a-b+b-c+c-a]-3[a-b+b-c][b-c+c-a][c-a+a-b]

⇒[0]-3[a-c][b-a][c-b]

⇒3(a-b)(c-a)(b-c)

⇒3(a-b)(b-c)(c-a)

(iii) Factors of (2a-3b)3+(3b-c)3+(c-2a)3 will be __

Sol :

⇒3(2a-3b)(3b-c)(c-2a)

(iv) Factors of (a+b+c)3+(a-b-c)3-8a3 will be __

[Hint: Given expression =(a+b+c)3+(a-b-c)3+(-2a)3]

Let x=a+b+c , y=a-b-c and z=-2a

Then , x+y+z=(a+b+c)+(a-b-c)-2a=0

⇒x3+y3+z3=3xyz

⇒(a+b+c)3+(a-b-c)3-8a3

⇒3(a+b+c)(a-b-c)(-2a)

⇒-6a(a+b+c)(a-b-c)

⇒6a(a+b+c)(b+c-a)

(v) If p=2-a, prove that a3+p3-8+6ap=0

[Hint: We have, p=2-a]

⇒a+p-2=0

Let x=a , y=p and z=(-2)

Then, x+y+z=0

⇒x3+y3+z3=3xyz

⇒(a)3+(p)3+(-2)3=3×a×p×(-2)

⇒a3+p3-8=-6ap or

a3+p3+6ap-8=0

Second method:

=a3+6ap+p3-8=a3+p3+(-2)3-3ap(-2)

={a+p+(-2)}{a2+p2+(-2)2-ap-p(-2)-a(-2)}

=(a+p-2)(a2+p2+4-ap+2p+2a)

=0×(a2+p2+4-ap+2p+2a)=0

#### Question 15

Factorize the following :

(i) x3-y3-z3-3xyz

Sol :

⇒(x-y-z)(x2+y2+z2+xy+xz-yz)

(ii) a3-b3-1-3ab

Sol :

⇒(a-b-1)(a2+b2+1+ab+a-b)

(iii) a3+b3-c3+3abc

Sol :

⇒(a+b-c)(a2+b2+c2-ab+bc+ca)

(iv) x3+8y3-z3+6xyz

Sol :

⇒(x+2y-z)(x2+4y2+z2-2xy+xz+2yz)

#### Question 16

Find value of x3-8y3-36xy-216, if x=2y+6

[Hint: Given, x-2y-6=0

Let a=x , b=-2y and c=-6

Then , a+b+c=0

⇒ a3+b3+c3=3abc

⇒x3+(-2y)3+(-6)3=3x(-2y)(-6)

⇒x3-8y3-216=36xy

⇒x3-8y3-36xy-216=0

]

Page 4.36

Second method :

Given , x-2y=6

⇒(x-2y)3=(6)3

⇒x3-3x2y(x-2y)-8y3=216

⇒x3-6xy(6)-8y3-216=0

⇒x3-36xy-8y3-216=0

#### Question 17

Find the value of a3+b3+c3-3abc, if

(i) a+b+c=14 and a2+b2+c2=68

[Hint: (i) a+b+c=14

⇒(a+b+c)2=196

⇒a2+b2+c2+2(ab+bc+ac)=196

⇒68+2(ab+bc+ac)=196

⇒ab+bc+ac

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

=14×(68-64)=56]

Sol :

⇒56

(ii) a+b+c=9 and ab+bc+ca=26

Sol:

⇒27

#### Question 18

Find the value of x3+y3+z3 , if x+y+z=11 ,

x2+y2+z2=45 and xyz=40

Sol :

⇒197

#### Question 19

Find the product

(i) (x+y-z)(x2+y2+z2-xy+yz+zx)

Sol :

⇒(x3+y3-z3+3xyz)

(ii) (3x-5y-4)(9x2+25y2+15xy+12x-20y+16)

[Hint: (i) We have (x+y-z)(x2+y2+z2-xy+yz+zx)]

=[x+y+(-z)][x2+y2+(-z)2-xy-y(-z)-x(-z)]

=(a+b+c)(a2+b2+c2-ab-bc-ac) , where a=x , b=y and c=-z

=a3+b3+c3-3abc=27x3-25y3-64-3(3x)(-5y)(-4)

=27x3-125y3-64-180xy

#### Question 20

(i) Find the value of x3+y3+z3 if x+y+z=1 ,

xy+yz+zx=-1 and xyz=-1

Sol :

⇒1

(ii) Simplify

[Hint: (i) We know that x3+y3+z3-3xyz

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

⇒x3+y3+z3-3xyz=(x+y+z)(x2+y2+z2+2xy+2yz+2zx-3xy-3yz-3zx)

[Adding and subtracting 2xy+2yz+2zx in second bracket only]

⇒x3+y3+z3=(x+y+z)(x+y+z)2-3(xy+yz+zx)]+3xyz

[Transposing 3xyz on R.H.S]

=1×[(1)2-3(-1)]+3(-1)

= 4-3 = 1

(ii) (a2-b2)+(b2-c2)+(c2-a2)=0

∴Numerator

=(a2-b2)+(b2-c2)+(c2-a2) =3(a2-b2)(b2-c2)(c2-a2)

=3(a-b)(a+b)(b-c)(b+c)(c-a)(c+a)..(i)

Again, (a-b)+(b-c)+(c-a)=0

∴(a-b)3+(b-c)3+(c-a)3=3(a-b)(b-c)(c-a)..(ii)

Hence, given expression=(a+b)(b+c)(c+a)

#### Question 21

Prove that:

(i) (a3+b3+c3)-3abc=1/2 (a+b+c){(a-b)2+(b-c)2+(c-a)2}

Sol :

(ii) (a+b)3+(b+c)3+(c+a)3-3(a+b)(b+c)(c+a) =2(a3+b3+c3-3abc)

(iii) Factorize : p3(q-r)3+q3(r-p)3+r3(p-q)3

Sol :

(iv) Find the value of (x-a)3+(x-b)3+(x-c)3=3(x-a)(x-b)(x-c) when a+b+c=3x

Sol :

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