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Question

Expand the following, Using suitable identities
(x23y)3

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Solution

(x2y3)3 is of the form (ab)3=a3b33ab(ab)

=x3(2y3)33×x×2y3(x2y3)

=x38y3272x2y+4xy23

=x38y327+4xy232x2y


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