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Question

Factorize 22a3+33b3+c336abc

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Solution

22a3+33b3+c336abc=(2a)3+(3b)3+(c)33×2a×3b×c=(2a+3b+c)[(2a)2+(3b)2+c22a×3b3b×cc×2a]=(2a+3b+c)(2a2+3b2+c26ab3bc2ca)


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