Question

Prove that (x+y)³+(y+z)³+(z+x)³-3(x+y×y+z×z+x)=2(x³+y³+z³-3xyz)

Answer 1

Let

$x+y=a \\ \\ y+z=b \\ \\ z+x=c$

$a-b=(x+y)-(y+z)=x+y-y-z=x-z \\ \\ b-c=(y+z)-(z+x)=y+z-z-x=y-x \\ \\ c-a=(z+x)-(x+y)=z+x-x-y=z-y$

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$(x+y)^3+(y+z)^3+(z+x)^3-3((x+y)(y+z)(z+x)) \\ \\ a^3+b^3+c^3-3abc \\ \\ (a+b+c)(a^2+b^2+c^2-ab-bc-ca) \\ \\ (x+y+y+z+z+x)(a^2+b^2+c^2-ab-bc-ca) \\ \\ 2(x+y+z)(a^2-ab+b^2-bc+c^2-ca) \\ \\ 2(x+y+z)(a(a-b)+b(b-c)+c(c-a)) \\ \\ 2(x+y+z)((x+y)(x-z)+(y+z)(y-x)+(z+x)(z-y)) \\ \\ 2(x+y+z)(x^2+yx-zx-yz+y^2+zy-xy-zx+z^2+xz-yz-xy) \\ \\ 2(x+y+z)(x^2+y^2+z^2-xy-yz-zx) \\ \\ 2(x^3+y^3+z^3-3xyz)$

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Hence proved!!!

Hope this helps. ^_^

This proof is on my own words, and not from any other sources.

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