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Question

Shew that (a2bc)33(a2bc)(b2ca)(c2ab)=(a3+b3+c33abc)2.

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Solution

Let X=a2bc,Y=b2ac,Z=c2ab
X+Y+Z=(a+wb+w2c)(a+w2b+wc)
X+wY+w2Z=(a+b+c)(a+wb+w2c)
X+w2Y+wZ=(a+b+c)(a+w2b+wc)
(X+Y+Z)(X+wY+w2Z)(X+w2Y+wZ)=X33XYZ
=(a+b+c)2(a+wb+w2c)2(a+w2b+wc)2
=(a3+b3+c33abc)2

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