Prove Nice Determinant Equations I often come across these kinds of problems in A-level exam papers:

Prove that 
  $$
\begin{vmatrix}
    (a+b)^2 & 1       & 1      \\
    a^2     & (1+b)^2 & a^2    \\
    b^2     & b^2     & (1+a)^2 
\end{vmatrix} 
= 2ab(1+a+b)^3
$$

or 

Prove that 
  $$
\begin{vmatrix}
    1  & 1  & 1 \\
    x  & y  & z \\
    yz & xz & xy 
\end{vmatrix} 
= (x-y)(y-z)(z-x)
$$

Questions which involve nice determinant results, but which are quite a pain to prove by expanding. 
Are there any tricks that one can use to prove such results? I'm familiar with Vandermonde matrices for example, but I haven't come across anything that might help me with these, especially the first one.
 A: By subtracting the first column from the second and the third we get that :
$$
\begin{split}
\begin{vmatrix}
    1  & 1  & 1 \\
    x  & y  & z \\
    yz & xz & xy 
\end{vmatrix}
&=\begin{vmatrix}
    1  & 0  & 0 \\
    x  & y-x  & z-x \\
    yz & z(x-y) & y(x-z) 
\end{vmatrix}\\
\\
&=(x-y)(z-x)\begin{vmatrix}
    1  & 0  & 0 \\
    x  & -1  & 1 \\
    yz & z & -y 
\end{vmatrix}\\
\\
&=(x-y)(z-x)\begin{vmatrix}
     -1  & 1 \\
     z & -y 
\end{vmatrix}\\
\\
& =(x-y)(y-z)(z-x)
\end{split}$$
A: $$
\begin{vmatrix}
    1  & 1  & 1 \\
    x  & y  & z \\
    yz & xz & xy 
\end{vmatrix} 
$$ it's a third degree cyclic homogeneous polynomial and its's obvious that
$$
\begin{vmatrix}
    1  & 1  & 1 \\
    x  & y  & z \\
    yz & xz & xy 
\end{vmatrix} 
= 0
$$ for $x=y$.
Id est,
$$
\begin{vmatrix}
    1  & 1  & 1 \\
    x  & y  & z \\
    yz & xz & xy 
\end{vmatrix} 
= A(x-y)(y-z)(z-x),
$$
where $A\in\mathbb Z$, and since the coefficient before $xy^2$ is equal to $1$, we obtain $A=1$.
By the same reasoning we can get a factor $ab(1+a+b)$ for the first determinant. 
It helps for the factorization after full expanding.
A: The idea is to use elementary row-column operations to have a simpler determinant. Since this is $3\times 3$, our goal is to get two successive zeros in a single row/column. 
You have
$$A=\begin{bmatrix}(a+b)^2 & 1 & 1 \\a^2 & (1+b)^2 & a^2 \\b^2 & b^2 & (1+a)^2 \end{bmatrix}$$
Then 
\begin{align}
\det A&=\det\begin{bmatrix}(a+b)^2-1 & 0 & 1 \\0 & (1+b)^2-a^2 & a^2 \\b^2-(1+a)^2 & b^2-(1+a)^2 & (1+a)^2 \end{bmatrix}\qquad\quad[C_j'=C_j-C_3\,,j=1,2]
\\\\&=\det\begin{bmatrix}(a+b+1)(a+b-1) & 0 & 1 \\0 & (a+b+1)(1+b-a) & a^2 \\(a+b+1)(b-1-a) & (a+b+1)(b-1-a) & (1+a)^2 \end{bmatrix}
\\\\&=(a+b+1)^2\det\begin{bmatrix}a+b-1 & 0 & 1 \\0 & 1+b-a & a^2 \\ b-1-a & b-1-a & (1+a)^2 \end{bmatrix}
\\\\&=(a+b+1)^2\det\begin{bmatrix}a+b-1 & 0 & 1 \\0 & 1+b-a & a^2 \\ -2a & -2 & 2a \end{bmatrix}\qquad\qquad[R_3'=R_3-(R_1+R_2)]
\\\\&=\frac{2(a+b+1)^2}{a}\det\begin{bmatrix}a+b-1 & 0 & 1 \\0 & a+ab-a^2 & a^2 \\ -a & -a & a \end{bmatrix}\qquad\qquad\quad[C_2'=aC_2]
\\\\&=2(a+b+1)^2\det\begin{bmatrix}a+b-1 & 0 & 1 \\0 & a+ab-a^2 & a^2 \\ -1 & -1 & 1 \end{bmatrix}
\\\\&=2(a+b+1)^2\det\begin{bmatrix}a+b & 1 & 1 \\a^2 & a+ab & a^2 \\ 0 & 0 & 1 \end{bmatrix}\qquad\qquad\qquad[C_j'=C_j+C_3\,,j=1,2]
\end{align}
Now expand with respect to the third row.
