$\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}{{a^2} + 1}&{ab}&{ac}\\{ab}&{{b^2} + 1}&{bc}\\{ca}&{cb}&{{c^2} + 1}\end{array}} \right|$

$= 1 + {a^2} + {b^2} + {c^2}.$

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

Applying ${C_1} \to {C_1} + {C_2} + {C_3}$ ,

we get
$L.H.S. = \left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}{1 + a(a + b + c)}&{ab}&{ac}\\{1 + b(a + b + c)}&{{b^2} + 1}&{bc}\\{1 + c(a + b + c)}&{cb}&{{c^2} + 1}\end{array}} \right|$

By property 5 and taking $(a + b + c)$common from ${C_1}$ in determinant II,

we get
L.H.S.: $= \left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}1&{ab}&{ac}\\1&{{b^2} + 1}&{bc}\\1&{cb}&{{c^2} + 1}\end{array}} \right| + (a + b + c)\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}a&{ab}&{ac}\\b&{{b^2} + 1}&{bc}\\c&{cb}&{{c^2} + 1}\end{array}} \right|$

Changing rows into columns,

we have
L.H.S.: $= \left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}1&1&1\\{ab}&{{b^2} + 1}&{bc}\\{ac}&{bc}&{{c^2} + 1}\end{array}} \right| + (a + b + c)\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}a&b&c\\{ab}&{{b^2} + 1}&{cb}\\{ac}&{bc}&{{c^2} + 1}\end{array}} \right|$

For determinant I we apply, ${C_1} \to {C_1} - {C_2},{C_2} \to {C_2} - {C_3}$

and
for determinant II we take out a common from ${C_1},$

we get
L.H.S.: $\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}0&0&1\\{ab - {b^2} - 1}&{{b^2} + 1 - cb}&{bc}\\{ac - bc}&{bc - {c^2} - 1}&{{c^2} + 1}\end{array}} \right|$ $+ a(a + b + c)\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}1&b&c\\b&{{b^2} + 1}&{bc}\\c&{bc}&{{c^2} + 1}\end{array}} \right|$

Applying ${C_2} \to {C_2} - b{C_1}$ and ${C_3} \to {C_3} - c{C_1}$ in determinant II,

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

Expanding along ${R_1},$

we have
L.H.S. $=$ $1[(ab - {b^2} - 1)(bc - {c^2} - 1) - (ac - bc)({b^2} + 1 - cb)] + a(a + b + c)$

$= [(a{b^2}c - ab{c^2} - ab - {b^3}c + {b^2}{c^2} + {b^2} - bc + {c^2} + 1 - (a{b^2}c + ac - ab{c^2} - {b^3}c - bc + {b^2}{c^2})] + a(a + b + c)$

$= a{b^2}c - ab{c^2} - ab - {b^3}c + {b^2}{c^2} + {b^2} - bc + {c^2} + 1 - a{b^2}c - ac + ab{c^2} + {b^3}c + bc - {b^2}{c^2} + a(a + b + c)$

$= - ab + {b^2} + {c^2} + 1 - ac + {a^2} + ab + ac$
$= 1 + {a^2} + {b^2} + {c^2} =$

R.H.S.
Hence, proved.