Question

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

Answer 1

$\begin{array}{c}\left|\begin{array}{ccc}a& b+c& a^2\\ b& c+a& b^2\\ c& a+b& c^2\end{array}\right|\\ ={\left|\begin{array}{ccc}a+b+c& b+c& a^2\\ a+b+c& c+a& b^2\\ a+b+c& a+b& c^2\end{array}\right|}^{C_1\to C_1+C_2}\end{array}$

Taking out common $a+b+c$ from $C_1$

$\begin{array}{c}=\left(a+b+c\right)\left|\begin{array}{ccc}1& b+a& a^2\\ 1& c+a& b^2\\ 1& a+b& c^2\end{array}\right|\\ =\left(a+b+c\right){\left|\begin{array}{ccc}1& b+c& a^2\\ 0& a-b& b^2-a^2\\ 0& a-c& c^2-a^2\end{array}\right|}^{R_2\to R_2-R_1{R}_3\to R_3-R_1}\\ =\left(a+b+c\right)\left|\begin{array}{ccc}1& b+c& a^2\\ 0& \left(a-b\right)& -\left(a+b\right)\left(a-b\right)\\ 0& \left(a-c\right)& -\left(a-c\right)\left(a+c\right)\end{array}\right|\end{array}$

Taking out common $(a-b)$ and $(a-c)$ from $R_1$ and $R_2$ respectively

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

Expanding along $C_1$

$=\left(a+b+c\right)\left(a-b\right)\left(a-c\right)\left[-a-c+a+b\right]$

$=\left(a+b+c\right)\left(a-b\right)\left(a-c\right)\left(b-c\right)$

$=-\left(a+b+c\right)\left(a-b\right)\left(b-c\right)\left(c-a\right)$

Proved