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

Performing the operations $C_1 \rightarrow C_1-bC_3$
and $C_2 \rightarrow C_2+aC_3$, I got
$$\Delta =\begin{vmatrix} 1+a^2+b^2&0&-2b \\ \ 0&1+a^2+b^2&2a \\\ b(1+a^2+b^2)&-a(1+a^2+b^2)&1-a^2-b^2 \end {vmatrix}$$
So I got the term $1+a^2+b^2$, but I am not able to pull it out. What should I do next?

$$\begin{aligned}\Delta &=\begin{vmatrix} 1+a^2+b^2&0&-2b \\ \ 0&1+a^2+b^2&2a \\\ b(1+a^2+b^2)&-a(1+a^2+b^2)&1-a^2-b^2 \end {vmatrix}\\
&=(1+a^2+b^2)\begin{vmatrix} 1&0&-2b \\ \ 0&1+a^2+b^2&2a \\\ b&-a(1+a^2+b^2)&1-a^2-b^2 \end {vmatrix}\\
&=(1+a^2+b^2)^2\begin{vmatrix} 1&0&-2b \\ \ 0& 1 &2a \\\ b &-a &1-a^2-b^2 \end {vmatrix}\\
&=(1+a^2+b^2)^2[(1-a^2-b^2 + 2a^2) + b(2b)]\\
&= (1+a^2+b^2)^3
\end{aligned}$$
As you can factor a coefficient in a column and then develop the determinant along the first column.
