Question

Prove that:

(x + y)³ + (y + z)³ + (z + x)³ − 3(x + y) (y + z) (z + x) = (x³ + y³ + z³ − 3xyz).

Answer 1

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
$$\begin{gathered} {\left(x+y\right)}^3+{\left(y+z\right)}^3+{\left(z+x\right)}^3-3\left(x+y\right)\left(y+z\right)\left(z+x\right) \\ =\left[\left(x+y\right)+\left(y+z\right)+\left(z+x\right)\right]\left[{\left(x+y\right)}^2+{\left(y+z\right)}^2+{\left(z+x\right)}^2-\left(x+y\right)\left(y+z\right)-\left(y+z\right)\left(z+x\right)-\left(z+x\right)\left(x+y\right)\right] \\ =2\left(x+y+z\right)\left[2\left(x^2+y^2+z^2+2xy+2yz+2zx\right)-xy-xz-y^2-yz-yz-xy-z^2-zx-xz-yz-x^2-xy\right] \\ =2\left(x+y+z\right)\left[\left(x^2+y^2+z^2+2xy+2yz+2zx\right)-xy-xz-yz\right] \\ =2\left(x+y+z\right)\left(x^2+y^2+z^2-xy-xz-yz\right) \end{gathered}$$
$=2\left(x^3+y^3+z^3-3xyz\right)$
Hence, (x + y)³ + (y + z)³ + (z + x)³ − 3(x + y) (y + z) (z + x) = 2(x³ + y³ + z³ − 3xyz).