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}
c&(a-b)&a \\
a&(b-c)&b \\
b&(c-a)&c \\
\end{matrix}\right|&= c\left|\begin{matrix}
(b-c)&b \\
(c-a)&c \\
\end{matrix}\right|-a\left|\begin{matrix}(a-b)&a\\(c-a)&c \\\end{matrix}\right|+b\left|\begin{matrix}(a-b)&a\\(b-c)&b \\\end{matrix}\right|\\
&=c[bc-c^2-bc+ab]-a[ac-bc-ac+a^2]+b[ab-b^2-ab+ac]\\
&=-c^3+abc+abc-a^3-b^3+abc\\
&=3abc-a^3-b^3-c^3.
\end{align*}
