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Question

Factorise using suitable identities:
27a3+8b327c3+18abc

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Solution

27a3+8b327c3+18abc is of the form x3+y3+z33xyz=(x+y+z)(x2+y2+z2xyyzzx)

(3a)3+(2b)3+(3c)33×3a×2b×3c

=(3a+2b3c)((3a)2+(2b)2+(3c)23a×2b2b×3c(3c)×3a)

=(3a+2b3c)(9a2+4b2+9c26ab+6bc+9ac)

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