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Question

# Prove that: (x + y)3 + (y + z)3 + (z + x)3 − 3(x + y) (y + z) (z + x) = (x3 + y3 + z3 − 3xyz).

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Solution

## Using the identities ${a}^{3}+{b}^{3}+{c}^{3}-3abc=\left(a+b+c\right)\left({a}^{2}+{b}^{2}+{c}^{2}-ab-bc-ca\right)\mathrm{and}{\left(a+b\right)}^{2}={a}^{2}+2ab+{b}^{2}$, we get ${\left(x+y\right)}^{3}+{\left(y+z\right)}^{3}+{\left(z+x\right)}^{3}-3\left(x+y\right)\left(y+z\right)\left(z+x\right)\phantom{\rule{0ex}{0ex}}=\left[\left(x+y\right)+\left(y+z\right)+\left(z+x\right)\right]\left[{\left(x+y\right)}^{2}+{\left(y+z\right)}^{2}+{\left(z+x\right)}^{2}-\left(x+y\right)\left(y+z\right)-\left(y+z\right)\left(z+x\right)-\left(z+x\right)\left(x+y\right)\right]\phantom{\rule{0ex}{0ex}}=2\left(x+y+z\right)\left[2\left({x}^{2}+{y}^{2}+{z}^{2}+2xy+2yz+2zx\right)-xy-xz-{y}^{2}-yz-yz-xy-{z}^{2}-zx-xz-yz-{x}^{2}-xy\right]\phantom{\rule{0ex}{0ex}}=2\left(x+y+z\right)\left[\left({x}^{2}+{y}^{2}+{z}^{2}+2xy+2yz+2zx\right)-xy-xz-yz\right]\phantom{\rule{0ex}{0ex}}=2\left(x+y+z\right)\left({x}^{2}+{y}^{2}+{z}^{2}-xy-xz-yz\right)$ $=2\left({x}^{3}+{y}^{3}+{z}^{3}-3xyz\right)$ Hence, (x + y)3 + (y + z)3 + (z + x)3 − 3(x + y) (y + z) (z + x) = 2(x3 + y3 + z3 − 3xyz).

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