Question

If x + y + z = 5 and xy + yz + zx = 7, then x³ + y³ + z³ – 3xyz = _________.

Answer 1

$$\begin{gathered} \mathrm{Given}: \\ x+y+z=5...\left(1\right) \\ xy+yz+zx=7...\left(2\right) \\ {\left(x+y+z\right)}^2=x^2+y^2+z^2+2xy+2yz+2zx \\ \Rightarrow 5^2=x^2+y^2+z^2+2\left(7\right) \\ \Rightarrow 25=x^2+y^2+z^2+14 \\ \Rightarrow x^2+y^2+z^2=11...\left(3\right) \\ \mathrm{Now}, \\ {x}^3+y^3+z^3-3xyz \\ =\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-zx\right)\left(\mathrm{Using}\mathrm{the}\mathrm{identity}:a^3+b^3+c^3-3abc=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)\right) \\ =\left(5\right)\left(11-7\right)\left(\mathrm{From}\left(1\right),\left(2\right)\mathrm{and}\left(3\right)\right) \\ =5\times 4 \\ =20 \end{gathered}$$

Hence, if x + y + z = 5 and xy + yz + zx = 7, then x³ + y³ + z³ – 3xyz = 20.