Question

Prove that $\left|\begin{array}{ccc}{a}^2& 2ab& b^2\\ b^2& a^2& 2ab\\ 2ab& b^2& a^2\end{array}\right|={\left(a^3+b^3\right)}^2$

Answer 1

$$\begin{gathered} \mathrm{Let}\mathrm{LHS}=∆=\left|a^{2}2ab{b}^2 \\ {b}^2{a}^{2}2ab \\ 2ab{b}^2{a}^2\right| \\ =a^2\left|a^{2}2ab \\ {b}^2{a}^2\right|-\left(2ab\right)\left|b^{2}2ab \\ 2ab{a}^2\right|+b^2\left|b^2{a}^2 \\ 2ab{b}^2\right|\left[\mathrm{Expanding}\right] \\ =a^2\left(a^4-2a{b}^3\right)-\left(2ab\right)\left(b^2{a}^2-4a^2{b}^2\right)+b^2\left(b^4-2a^{3}b\right) \\ =a^6-2a^3{b}^3-2a^3{b}^3+8a^3{b}^3+b^6-2a^3{b}^3 \\ =a^6+2a^3{b}^3+b^6 \\ ={\left(a^3\right)}^2+2a^3{b}^3+{\left(b^3\right)}^2 \\ ={\left(a^3+b^3\right)}^2 \\ =\mathrm{RHS} \end{gathered}$$

Hence proved.