Without using the Rule of Sarrus, prove that: Without using the Rule of Sarrus, prove that:
$$\left|
\begin{matrix}
(b+c)&(a-b)&a \\
(c+a)&(b-c)&b \\
(a+b)&(c-a)&c \\
\end{matrix}\right|=3abc-a^3-b^3-c^3$$
My Approach:
$$LHS=
 \left|
\begin{matrix}
(b+c)&(a-b)&a \\
(c+a)&(b-c)&b \\
(a+b)&(c-a)&c \\
\end{matrix}\right|$$
$$C_1\to C_1+C_2$$
$$=
\left|
\begin{matrix}
(c+a)&(a-b)&a \\
(a+b)&(b-c)&b \\
(b+c)&(c-a)&c \\
\end{matrix}\right|$$
$$C_1\to C_1-C_3$$
$$=
\left|
\begin{matrix}
c&(a-b)&a \\
a&(b-c)&b \\
b&(c-a)&c \\
\end{matrix}\right|$$
How do I complete the rest?

\begin{align}
 \left|
\begin{matrix}
(b+c)&(a-b)&a \\
(c+a)&(b-c)&b \\
(a+b)&(c-a)&c \\
\end{matrix}\right|&=
 \left|
\begin{matrix}
a+b+c&a-b&a \\
a+b+c&b-c&b \\
a+b+c&c-a&c \\
\end{matrix}\right|\\
&= (a+b+c)\left|
\begin{matrix}
1&a-b&a \\
1&b-c&b \\
1&c-a&c \\
\end{matrix}\right|\\
&= (a+b+c)\left|
\begin{matrix}
1&-b&a \\
1&-c&b \\
1&-a&c \\
\end{matrix}\right|\end{align}
Can you continue from here?
