Question

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

Answer 1

$\left|\begin{array}{ccc}a& a^2& b+c\\ b& b^2& c+a\\ c& c^2& a+b\end{array}\right|$

Using operation, $C_1\Rightarrow C_1+C_3$

$=\left|\begin{array}{ccc}a+b+b& a^2& b+c\\ a+b+c& b^2& c+a\\ a+b+c& c^2& a+b\end{array}\right|$

Taking out $(a+b+c)$ as common term, we get

$=(a+b+c)\left|\begin{array}{ccc}1& a^2& b+c\\ 1& b^2& c+a\\ 1& c^2& a+b\end{array}\right|$

Using operation, $R_1\Rightarrow R_1-R_2,R_2\Rightarrow R_2-R_3$

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

Taking out $(a-b)$ and $(b-c)$ as common from $R_1$ and $R_2$ respectively, we get

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

Now, expanding along the element in $R_3$ and $C_1$, we get

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