Question

Prove the following :
$\left|\begin{array}{ccc}2ab& a^2& b^2\\ a^2& b^2& 2ab\\ b^2& 2ab& a^2\end{array}\right|=-{(a^3+b^3)}^2.$

Answer 1

Apply $C_1+C_2+C_3$ and take out ${(a+b)}^2$ common from $C_1$
$△={(a+b)}^2\left|\begin{array}{ccc}1& a^2& b^2\\ 1& b^2& 2ab\\ 1& 2ab& a^2\end{array}\right|$
Apply $R_2-R_1$; $R_3-R_1$ to make two zeros.
$△={(a+b)}^2\left|\begin{array}{ccc}1& a^2& b^2\\ 0& b^2-a^2& b(2a-b)\\ 0& a(2b-a)& a^2-b^2\end{array}\right|$
$={(a+b)}^2\left|\begin{array}{cc}{b}^2-a^2& b(2a-b)\\ a(2b-a)& a^2-b^2\end{array}\right|$
$={(a+b)}^2[-{(a^2-b^2)}^2-ab(2b-a)(2a-b)]$
$={-(a+b)}^2[{(a^2-b^2)}^2+4a^2{b}^2-2ab(a^2+b^2)+a^2{b}^2]$
$={-(a+b)}^2[{(a^2-b^2)}^2-2ab(a^2+b^2)+a^2{b}^2]$
$={-(a+b)}^2{[a^2+b^2-ab]}^2=-{(a3+b^3)}^2$