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Question

If x+y+z=0, show that x3+y3+z3=3xyz.


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Solution

Prove the expression

It is given that x+y+z=0

We know that,

x3+y3+z33xyz=(x+y+z)(x2+y2+z2xyyzxz)x3+y3+z33xyz=(0)(x2+y2+z2xyyzxz)x+y+z=0x3+y3+z33xyz=0x3+y3+z3=3xyz

Hence, x3+y3+z3=3xyz.


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