Prove that $\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}{{a^2}}&{bc}&{ac + {c^2}}\\{{a^2} + bc}&{{b^2}}&{ac}\\{ab}&{{b^2} + bc}&{{c^2}}\end{array}} \right| = 4{a^2}{b^2}{c^2}.$

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

Taking $a,b,c$common from ${C_1},{C_2}\& {C_3}$,

we get
$= abc\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}a&c&{a + c}\\{a + b}&b&a\\b&{b + c}&c\end{array}} \right|$

Applying ${C_3} \to {C_3} - ({C_1} + {C_2})$ ,

we get
$= abc\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}a&c&0\\{a + b}&b&{ - 2b}\\b&{b + c}&{ - 2b}\end{array}} \right|$

Taking$- 2b$ common from ${C_3},$

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

Expanding the determinant along ${R_1}$,

we get
$= - 2b(abc)[a(b - b - c) - c(a + b - c)].$

$= - 2b(abc)[ - 2ac] = 4{a^2}{b^2}{c^2} = R.H.S$

Hence, proved.