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?

you can expand the matrix determinant and look for "shortcuts" , my solution : 
$$
\begin{vmatrix} 
x & a & b \\
x^2 & a^2 & b^2 \\
a+b & x+b & x+a \\
\end{vmatrix} =
x \begin{vmatrix} 
x & a\\
x^2 & a^2\\
\end{vmatrix}-a\begin{vmatrix} 
x^2 & b^2\\
a+b & x+a\\
\end{vmatrix}+b\begin{vmatrix} 
x^2 & a^2\\
a+b & x+b\\
\end{vmatrix}
$$
now just expand the matrices determinant:
$$=x(x^2a^2-ax^2)-a(x^2(x+a)-b^2(a+b))+b(x^2(x+b)-a^2(a+b))=$$
open the brackets:
$$x^3a^2-ax^3-ax^3-a^2x^3-a^2b^2+ab^3+bx^3+x^2b^2-a^3b-a^2b^2=
$$
eliminate opposite sign expressions :
$$
a^3x-b^3x-ax^3+ab^3+bx^3-a^3b=
$$
$$x(a^3-b^3)-x^3(a-b)-ab(a^2-b^3)=$$
extract common divisor : 
$$[b-a](x^3-x(a^2+b^2+ab)+ab(a+b))=$$
$$[b-a](x^3-x((a+b)^2-ab)+ab(a+b))=$$
$$[b-a](x(x^2-(a+b)^2)+ab(x+a+b))=[b-a][x+a+b](x(x-a-b)+ab)=[b-a][x+a+b](x^2-x(a+b)+ab)=[b-a][x+a+b][x-a][x-b].
$$
$$
 \square
$$

