Question

If x + y + z = 0, then (x + y)³ + (y + z)³ + (z + x)³ = ________.

Answer 1

$$\begin{gathered} \mathrm{Given}: \\ x+y+z=0 \\ \Rightarrow x+y=-z...\left(1\right) \\ \Rightarrow z+y=-x...\left(2\right) \\ \Rightarrow x+z=-y...\left(3\right) \\ \mathrm{Now}, \\ {\left(x+y\right)}^3+{\left(y+z\right)}^3+{\left(z+x\right)}^3 \\ ={\left(-z\right)}^3+{\left(-x\right)}^3+{\left(-y\right)}^3\left(\mathrm{From}\left(1\right),\left(2\right)\mathrm{and}\left(3\right)\right) \\ =-z^3-x^3-y^3 \\ =-\left(z^3+x^3+y^3\right) \\ =-3xyz\left(\mathrm{Using}\mathrm{the}\mathrm{identity}:\mathrm{if}\left(a+b+c\right)=0\mathrm{then}{a}^3+b^3+c^3=3abc\right) \end{gathered}$$

Hence, if x + y + z = 0, then (x + y)³ + (y + z)³ + (z + x)³ = −3xyz.