Question

If (a + b + c) {(a – b)² + (b – c)² + (c – a)²} = k(a³ + b³ + c³ – 3abc), then k = ________.

Answer 1

$$\begin{gathered} \mathrm{Given}: \\ \left(a+b+c\right)\left\{{\left(a-b\right)}^2+{\left(b-c\right)}^2+{\left(c-a\right)}^2\right\}=k\left(a^3+b^3+c^3-3abc\right) \\ \left(a+b+c\right)\left\{{\left(a-b\right)}^2+{\left(b-c\right)}^2+{\left(c-a\right)}^2\right\}=k\left(a^3+b^3+c^3-3abc\right) \\ \Rightarrow \left(a+b+c\right)\left\{a^2+b^2-2ab+b^2+c^2-2bc+c^2+a^2-2ac\right\}=k\left(a^3+b^3+c^3-3abc\right)\left(\mathrm{Using}\mathrm{the}\mathrm{identity}:{\left(x-y\right)}^2=x^2+y^2-2xy\right) \\ \Rightarrow \left(a+b+c\right)\left\{2a^2+2b^2+2c^2-2ab-2bc-2ac\right\}=k\left(a^3+b^3+c^3-3abc\right) \\ \Rightarrow \left(a+b+c\right)\left\{2a^2+2b^2+2c^2-2ab-2bc-2ac\right\}=k\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)\left(\mathrm{Using}\mathrm{the}\mathrm{identity}:a^3+b^3+c^3-3abc=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)\right) \\ \Rightarrow 2\left(a+b+c\right)\left\{a^2+b^2+c^2-ab-bc-ac\right\}=k\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right) \\ \Rightarrow k=2 \end{gathered}$$

Hence, if (a + b + c) {(a – b)² + (b – c)² + (c – a)²} = k(a³ + b³ + c³ – 3abc), then k = 2.