By using properties of determinants, show that:

$\Delta=\left|\begin{array}{cll}1+a^{2}-b^{2} & 2 a b & -2 b \\ 2 a b & 1-a^{2}+b^{2} & 2 a \\ 2 b & -2 a & 1-a^{2}-b^{2}\end{array}\right|$

Applying $R_{1} \rightarrow R_{1}+b R_{3}$ and $R_{2} \rightarrow R_{2}-a R_{2}$, we have:

$\Delta=\left|\begin{array}{lll}1+a^{2}+b^{2} & 0 & -b\left(1+a^{2}+b^{2}\right) \\ 0 & 1+a^{2}+b^{2} & a\left(1+a^{2}+b^{2}\right) \\ 2 b & -2 a & 1-a^{2}-b^{2}\end{array}\right|$

$=\left(1+a^{2}+b^{2}\right)^{2}\left|\begin{array}{lll}1 & 0 & -b \\ 0 & 1 & a \\ 2 b & -2 a & 1-a^{2}-b^{2}\end{array}\right|$

Expanding along R1, we have:

$\begin{aligned} \Delta &=\left(1+a^{2}+b^{2}\right)^{2}\left[(1)\left|\begin{array}{ll}1 & a \\ -2 a & 1-a^{2}-b^{2}\end{array}\right|-b\left|\begin{array}{ll}0 & 1 \\ 2 b & -2 a\end{array}\right|\right] \\ &=\left(1+a^{2}+b^{2}\right)^{2}\left[1-a^{2}-b^{2}+2 a^{2}-b(-2 b)\right] \\ &=\left(1+a^{2}+b^{2}\right)^{2}\left(1+a^{2}+b^{2}\right) \\ &=\left(1+a^{2}+b^{2}\right)^{3} \end{aligned}$