Question

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

Answer 1

$L.H.S.=\left|\begin{array}{ccc}a& b& c\\ a^2& b^2& c^2\\ bc& ca& ab\end{array}\right|$ = $\left|\begin{array}{ccc}a-c& b-c& c\\ a^2-c^2& b^2-c^2& c^2\\ bc-ab& ca-ab& ab\end{array}\right|$ $(C_1\to C_1-C_3,C_2\to C_2-C_3)$

$=(a-c)(b-c)\left|\begin{array}{ccc}1& 1& c\\ a+c& b+c& c^2\\ -b& -a& ab\end{array}\right|$ (Taking (a-b) and (b-c) common from column 1 and column 2 respectively)

$=(a-c)(b-c)\left|\begin{array}{ccc}0& 1& c\\ a-b& b+c& c^2\\ a-b& -a& ab\end{array}\right|$ $(C_1\to C_1-C_2)$

$=(a-b)(b-c)(a-c)\left|\begin{array}{ccc}0& 1& c\\ 1& b+c& c^2\\ 1& -a& ab\end{array}\right|$ (Taking (a-b) common from column 1)

$=(a-b)(b-c)(a-c)\left|\begin{array}{ccc}0& 1& c\\ 0& a+b+c& c^2-ab\\ 1& -a& ab\end{array}\right|$ $(R_2\to R_2-R_3)$

$=(a-b)(b-c)(a-c)[c^2-ab-cb-c^2-ac]$

$=(a-b)(b-c)(c-a)(ab+bc+ca)=R.H.S.$