Finding the Determinant of a $3\times 3$ matrix. Show that:
$$
\begin{vmatrix} 
x & a & b \\
x^2 & a^2 & b^2 \\
a+b & x+b & x+a \\
\end{vmatrix} = (b - a)(x-a)(x-b)(x+a+b)
$$
I tried expanding the whole matrix out, but it looks like a total mess. Does anyone have an idea how this could be simplified?

Regard the matrix as a polynomial $f(x)$ in $x$.
Note that $a$ and $b$ are roots, because when you set $x = a$ you get that two of the columns are equal, and ditto for $x = b$.
Note that $- a - b$ is a root, because when you set $x = -a-b$ you get that two of the rows are equal.
So
$$
f(x) = c (x - a) (x - b) (x + a + b)
$$
for some constant $c$. 
To compute $c$, note that you obtain $x^{3}$ only from the following term of the expansion with respect to the first column
$$
x^{2} \cdot \begin{vmatrix}a&b\\x+b&x+a\end{vmatrix}.
$$
