Factorise the determinant $\det\Bigl(\begin{smallmatrix} a^3+a^2 & a & 1 \\ b^3+b^2 & b & 1 \\ c^3+c^2 & c &1\end{smallmatrix}\Bigr)$ Factorise the determinant $\det\begin{pmatrix} a^3+a^2 & a & 1 \\ b^3+b^2 & b & 1 \\ c^3+c^2 & c &1\end{pmatrix}$.
My textbook only provides two simple examples.
Really have no idea how to do this type of questions..

$$\begin{align*}
\det\begin{pmatrix} a^3+a^2 & a & 1 \\ b^3+b^2 & b & 1 \\ c^3+c^2 & c &1\end{pmatrix} & = \det\begin{pmatrix} a^2 & a & 1 \\ b^2 & b & 1 \\ c^2 & c &1\end{pmatrix}+\det\begin{pmatrix} a^3 & a & 1 \\ b^3 & b & 1 \\ c^3 & c &1\end{pmatrix}\\
 & = -(a-b)(b-c)(c-a) -(a-b)(b-c)(c-a)(a+b+c)\\
 &= (a-b)(b-c)(a-c)(a+b+c+1)
\end{align*}$$
see Vandermonde matrix

Hey guys, editing my post to address concerns. You're right, I didn't put in all the details because I thought the method was clear once you've seen Vandermonde Determinants. Here it is explicitly:
$$\begin{align*}
\det\begin{pmatrix} a^3 & a & 1 \\ b^3 & b & 1 \\ c^3 & c &1\end{pmatrix} & = \det\begin{pmatrix} a^3+(b+c)a^2 & a & 1 \\ b^3 +(a+c)b^2& b & 1 \\ c^3+(a+b)c^2 & c &1\end{pmatrix}\\
& = \sum_{cyc}\det\begin{pmatrix} a^3 & a & 1 \\ ab^2 & b & 1 \\ ac^2 & c &1\end{pmatrix} &  \\
&= \sum_{cyc}a\det\begin{pmatrix} a^2 & a & 1 \\ b^2 & b & 1 \\ c^2 & c &1\end{pmatrix} 
\end{align*}$$
(At this point we have shown the result.)
Okay, so I guessed intuitively that $$\det\begin{pmatrix} (b+c)a^2 & a & 1 \\ (a+c)b^2& b & 1 \\ (a+b)c^2 & c &1\end{pmatrix}=0$$But its easy to prove, by finding the eigenvector:
$$\begin{pmatrix}1 \\ -(ab+bc+ca) \\ abc\end{pmatrix}$$
