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

L.H.S. $= \left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}{1 + {a^2} - {b^2}}&{2ab}&{ - 2b}\\{2ab}&{1 - {a^2} + {b^2}}&{2a}\\{2b}&{ - 2a}&{1 - {a^2} - {b^2}}\end{array}} \right|$

Applying ${C_1} \to {C_1} - b{C_3}$ and${C_2} \to {C_2} + a{C_3}$,

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

Taking $(1 + {a^2} + {b^2})$common from ${C_1}$ and${C_2}$,

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

Applying ${R_3} \to {R_3} - b{R_1},$

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

Expanding along ${C_1},$

we get
$= {(1 + {a^2} + {b^2})^2}[1((1 - {a^2} + {b^2} + 2{a^2})]$
$= {(1 + {a^2} + {b^2})^2}(1 + {a^2} + {b^2})$

$= {(1 + {a^2} + {b^2})^3}$ $=$ R.H.S.
Hence, proved.