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Question

Question 31
Expand the following:
(i) (3a2b)3

(ii) (1x+y3)3

(iii) (413x)3

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Solution

(i) (3a2b)3=(3a)3+(2b)3+3(3a)(2b)(3a2b)
[Using the identity, (ab)3=a3b3+3a(b)(ab)]
=27a38b318ab(3a2b)
=27a38b354a2b+36ab2
=27a354a2b+36ab28b3

(ii) (1x+y3)3=(1x)3+(y3)3+3(1x)(y3)(1x+y3)
[Using the identity, (a+b)3=a3+b3+3ab(a+b)]
=1x3+y327+yx(1x+y3)=1x3+y327+yx2+y23x

(iii) (413x)3=(4)3+(13x)3+3(4)(13x)(413x)
[Using the identity, (ab)3=a3b3+3a(b)(ab)]
=64127x34x(413x)
=64127x316x+43x2

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