Question
By using properties of determinants, show that:
$\begin{vmatrix}1+a^2-b^2&2ab&-2b\\2ab&1-a^2+b^2&2a\\2b&-2a&1-a^2-b^2\end{vmatrix}=(1+a^2+b^2)^3$

Answer

$\text{L.H.S.}=\begin{vmatrix}1+a^2-b^2&2ab&-2b\\2ab&1-a^2+b^2&2a\\2b&-2a&1-a^2-b^2\end{vmatrix}$
$=\begin{vmatrix}1+a^2+b^2&0&-2b\\0&1+a^2+b^2&2a\\b(1+a^2+b^2)&-a(1+a^2+b^2)&1-a^2-b^2\end{vmatrix}\ \left[\text{C}_1\rightarrow\text{C}_1-b\ \text{C}_3\ \text{and C}_2\rightarrow\text{C}_2+a\ \text{C}_3\right]$

$=(1+a^2+b^2)^2\begin{vmatrix}1&0&-2b\\0&1&2a\\b&-a&1-a^2-b^2\end{vmatrix}$

$=(1+a^2+b^2)^2\begin{vmatrix}1&0&-2b\\0&1&2a\\0&-a&1-a^2+b^2\end{vmatrix}\ \left[\text{R}_3\rightarrow\text{R}_3-b\text{R}_1\right]$

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

$=(1+a^2+b^2)^2(1-a^2+b^2+2a^2)=(1+a^2+b^2)^3=\text{R.H.S.}$

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