Prove the following determinant Prove the following:
$$\left|
\begin{matrix}
(b+c)&a&a \\
b&(c+a)&b \\
c&c&(a+b) \\
\end{matrix}\right|=4abc$$
My Attempt:
$$\left|
\begin{matrix}
(b+c)&a&a \\
b&(c+a)&b \\
c&c&(a+b) \\
\end{matrix}\right|$$
Using $R_1\to R_1+R_2+R_3$
$$\left |
\begin{matrix}
2(b+c)&2(a+c)&2(a+b) \\
b&(c+a)&b \\
c&c&(a+b) \\
\end{matrix}\right|$$
Taking common $2$ from $R_1$
 $$2\left|
\begin{matrix}
(b+c)&(a+c)&(a+b) \\
b&(c+a)&b \\
c&c&(a+b) \\
\end{matrix}\right|$$
How do I proceed further?

The determinant of the given matrix is a symmetric polynomial with degree $3$ in the variables $a,b,c$. By the Laplace expansion the determinant equals zero if $a,b$ or $c$ equal zero, hence the determinant is a constant multiple of $abc$. In order to find which multiple, it is enough to evaluate the determinant at $(a,b,c)=(1,2,3)$, for instance:
$$\det\begin{pmatrix}5 & 1 & 1\\ 2 & 4 & 2 \\ 3 & 3 & 3 \end{pmatrix}=6\det\begin{pmatrix}5 & 1 & 1\\ 1 & 2 & 1 \\ 1 & 1 & 1 \end{pmatrix}=6\det\begin{pmatrix}4 & 0 & 0\\ 0 & 1 & 0 \\ 1 & 1 & 1 \end{pmatrix}=24.$$
This leads to $\det M(a,b,c) = 4abc$ as wanted.
