Finding the missing entry in determinant We have, 
$\begin{vmatrix}
a^2+2a & 2a+1 & 1 \\ 
2a+1 & x & 1\\ 
3 & 3 & 1
\end{vmatrix}=(a-1)^3$ I am asked to prove $x=a+2$ by using properties of determinants.
I've no idea how to solve it. Though tried to add or substract one column or row with others. No combinations had worked.

$$\begin{vmatrix}
a^2+2a & 2a+1 & 1 \\ 
2a+1 & x & 1\\ 
3 & 3 & 1
\end{vmatrix}\underbrace{=}_{C_1-3C_3, C_2-3C_3} \begin{vmatrix}
a^2+2a-3 & 2a-2 & 1 \\ 
2a-2 & x-3 & 1\\ 
0 & 0 & 1
\end{vmatrix}=\begin{vmatrix}
a^2+2a-3 & 2a-2 \\ 
2a-2 & x-3\\ 
\end{vmatrix}$$
$$=(x-3)(a^2+2a-3)-4(a-1)^2=(a-1)^3 \iff x-3=\frac{(a-1)^3+4(a-1)^2}{a^2+2a-3}=a-1,$$ from where $x=a+2.$
