SChand CLASS 9 Chapter 4 Factorisation Exercise 4(H)

 Exercise 4(H)

Question 1

a³ + 1

Sol :

=x3+1

=(x)3+(1)3

=(x+1)(x2x+1)


Question 2

x³ + 8

Sol :

=x3+8
=(x)3+(2)3
=(x+2)[(x)2x×2+(2)2]
=(x+2)(x22x+4)

Question 3

8x³ + 1

Sol :

=8x3+1
=(2x)3+(1)3
=(2x+2)[(2x)22x×2+(1)2]
=(2x+1)(4x22x+1)

Question 4

x³ – 27

Sol :

=x327
=(x)3(3)3
=(x3)[(x)2+x×3+(3)2]
=(x3)(x2+3x+9)

Question 5

a³ – 8

Sol :

=a38

=(a)3(2)3

=(a2)[(a)2+a×2+(2)2]

=(a2)(a2+2a+4)


Question 6

27m³ – 8

Sol :

=27m38
=(3m)3(2)3
=(3m2)[(3m)2+3m×2+(2)2]
=(3m2)[(9m)2+6m+4]

Question 7

x³+ 64

Sol :

=x3+64=(x3)+(4)3=(x+4)[(x)2x×4+(4)2]=(x+4)(x24x+16)


Question 8

8a3b6

Sol :

=8a3b6=(2a)3(b2)3=(2ab2)[(2a)2+2a×b2+(b2)2]=(2ab2)(4a2+2ab2+b4]


Question 9

x6+8b3

Sol :

=x6+8b3=(x2)3+(2b)3=(x2+2b)[(x2)2x2×2b+(2b)2]=(x2+2b)(x42x2b+4b2)


Question 10

8a³ + 21b³

Sol :

=8a3+27b3=(2a)3+(3b)3=(2a+3b)[(2a)32a×3b+(3b)2]=(2a+3b)(4a26ab+9b2)


Question 11

27x³ – 8y³

Sol :

=27x38y3
=(3x)3(2y)3
=(3x2y)[(3x)2+3x×2y+(2y)2]
=(3x2y)(9x2+6xy+4y2)

Question 12

128x³ + 2

Sol :

=128x3+2=2(64x3+1)=2[(4x)3+(1)3]=2(4x+1)[(4x)24x×1+(1)2]=2[4x+2)(16x24x+1)


Question 13

Factorise : 

8x3127y3

Sol :

=8x3127y3

=(2x)3(13y)3
=(2x13y)[(2x)2+2x×13y+(13y)2]
=(2x19y)(4x2+23xy+19y2)


Question 14

343x3y+512y4

Sol :

=343x3y+512y4

=y(343x3+512y3)

=y[(7x)3+(8y)3]

=y[(7x+8y)[(7x)27x×8y+(8y)2]]

=y(7x+8y)(49x256xy+64y2)



Question 15

(2a + b)³ + (a + 2b)³

Sol :

=(2ab+b)3+(a+2b)3
Let 2a+b=x and a+2b=y

(x3)+(y)3

=(x+y)(x2xy+y2)

=(2a+b)3+(a+2b)3=(2a+b+2b)

=[(2a+b)2(2a+b)(a+2b)+(a+2b)2]

=(3a+3b)(4a2+4ab+b22a24abab2b2+a2+4b2+4ab
=3(a+b)[3a2+3ab+3b2]
=3(a+b)3(a2+ab+b2)
=9(a+b)(a2+ab+b2)


Question 16

27 (m + 2n)³ + (m – 6n)³

Sol :

=27(m+2n)3+(m6n)3
Let m+2n=a and m-6n=b

27a3+b3=(3a)3+(b)3

=(3a+b)[(3a)23ab+(b)2]

=(3a+b)(9a23ab+b2)

So, 27(m+2n)3+(m6n)3=[3(m+2n)+m6n][9(m2n)23(m+2n)(m6n)+(m6n)2
=[3m+6n+m6n][9(m2+4n2+4mn)3(m26mn+2mn12n2)+m2+36n212mn]
=4m[9m2+36n2+36mn3m2+18mn6mn+36n2+m2+36n212mn]
=4m[7m2+36mn+408n2]

Question 17

8 (a + b)³ – 21c³

Sol :

=8(a+b)327(c)3
=[2(a+b)]3(3c)3
=[2(a+b)3c][{2(a+b}2+2(a+b)3c+(3c)2]
=(2a+2b3c)[4(a2+b2+2ab)+6ac+6bc+9c2]
=(2a+2b3c)[4a2+4b2+8ab+6ac+6bc+9c2]
=(2a+2b3c)(4a2+4b2+9c2+8ab+6ac+6bc)


Question 18

x61

Sol :

=x61

=(x3)2(1)2

=(x3+1)(x31)

=[(x)3+(1)2][(x)3(1)3]

=(x+1)(x2x+1)(x1)(x2+x+1)

=(x+1)(x1)(x2x+1)(x2+x+1)



Question 19

64a6b6

Sol :

=64a6b6
=(8a3)2(b3)2

=(8a3+b3)(8a3b3)

=[(2a)3+(b)3][(2a)3(b)3]

=(2a+b)[(2a)22a×b+(b)2](2ab)[(2a)2+2a×b+b2]

=(2a+b)(4a22ab+b2)(2ab)(4a2+2ab+b2)

=(2a+b)(2ab)(4a2+2ab+b2)(4a22ab+b2)


Question 20

a³ – b³ + 4 (a – b)

Sol :

=a3b3+4(ab)=(ab)(a2+ab+b2)+u(ab)=(ab)(a2+ab+b3+4)


Question 21

x31x36x+6x

Sol :

=x31x36x+6x=(x1x)(x2+1+1x2)6(x1x)=(x1x)(x2+1+1x26)=(x1x)(x25+1x2)


Question 22

64a³ + 125b³ + 12a²b + 15ab²

Sol :

=64a3+125b3+12a2b+15ab2=[(4a)3+(5b)3]+3ab(4a+5b)=(4a+5b)[(4a)24a×5b+(5b)2]+3ab(4a+5b)

=(4a+5b)(16a220ab+25b2)+3ab(4a+5b)

=(4a+5b)(16a220ab+25b2+3ab)

=(4a+sb)(16a217ab+25b2)


Question 23

375 (a – b)³+ 3

Sol :

=375(ab)3+3

=3[125(ab)3+1]

=3[{5(ab)y3+(1)3]

=3[{5(ab)+1}{(5(ab)}25(ab×1+(1)2]

=3[(5a5b+1)(25(a2+b22ab)5a+5b+1)]

=3(5a5b+1)(25b250ab5a+5b+1)


Question 24

a4+b4=a2b2 , Show that a6+b6=0

Sol :

a4+b4=a2b2

L.H.S

=a6+b6=(a2)3+(b2)3

=(a2+b2)[a4a2b2+b4]

=(a2+b2)(a4+b4a2b2)

=(a2+b2)(a2b2a2b2)

a4+b4=a2b2 already given

=(a2+b2)×0=0

R.H.S

L.H.S=R.H.S

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