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?

Here is a way to break it down to a factor and only one $2\times 2$-determinant containing only binomials before expanding:
$$\left|
\begin{matrix}
(b+c)&(a-b)&a \\
(c+a)&(b-c)&b \\
(a+b)&(c-a)&c \\
\end{matrix}\right|\stackrel{R_3 \mapsto R_3+R_2+R_1}{=}
\left|
\begin{matrix}
(b+c)&(a-b)&a \\
(c+a)&(b-c)&b \\
2(a+b+c)&0&(a+b+c) \\
\end{matrix}\right| =
(a+b+c)\left|
\begin{matrix}
(b+c)&(a-b)&a \\
(c+a)&(b-c)&b \\
2&0&1 \\
\end{matrix}\right|
\stackrel{C_1 \mapsto C_1-2C_3}{=}
(a+b+c)\left|
\begin{matrix}
(b+c-2a)&(a-b)&a \\
(c+a-2b)&(b-c)&b \\
0&0&1 \\
\end{matrix}\right|
\stackrel{C_1 \mapsto C_1+C_2}{=}
(a+b+c)\left|
\begin{matrix}
(c-a)&(a-b)\\
(a-b)&(b-c)
\end{matrix}\right| =$$ $$=
(a+b+c)(ab+ac+bc-a^2-b^2-c^2) = 3abc - (a^3 + b^3 + c^3)
$$
