Question

If a + b + c = 5 and ab + bc + ca = 10, prove that a³ + b³ + c³ − 3abc = −25.

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{have}: \\ {a}^2+b^2+c^2={\left(a+b+c\right)}^2-2\left(ab+bc+ca\right) \end{gathered}$$
$$\begin{gathered} =5^2-\left(2\times 10\right) \\ =25-20 \\ =5 \end{gathered}$$
Now,
$$\begin{gathered} a^3+b^3+c^3-3abc=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right) \\ =\left(a+b+c\right)\left[a^2+b^2+c^2-\left(ab+bc+ca\right)\right] \end{gathered}$$
$$\begin{gathered} =5\times \left(5-10\right) \\ =5\times \left(-5\right) \\ =-25 \end{gathered}$$