Prove that $\left| {\begin{array}{llllllllllllllllllll}{bc - {a^2}}&{ca - {b^2}}&{ab - {c^2}}\\{ca - {b^2}}&{ab - {c^2}}&{bc - {a^2}}\\{ab - {c^2}}&{bc - {a^2}}&{ca - {b^2}}\end{array}} \right|$ is divisible by $(a + b + c)$ and find the quotient.

Let $\Delta = \left| {\begin{array}{cccccccccccccccccccc}{bc - {a^2}}&{ca - {b^2}}&{ab - {c^2}}\\{ca - {b^2}}&{ab - {c^2}}&{bc - {a^2}}\\{ab - {c^2}}&{bc - {a^2}}&{ca - {b^2}}\end{array}} \right|$

$= \left| {\begin{array}{llllllllllllllllllll}{bc - {a^2} - ca + {b^2}}&{ca - {b^2} - ab + {c^2}}&{ab - {c^2}}\\{ca - {b^2} - ab + {c^2}}&{ab - {c^2} - bc + {a^2}}&{bc - {a^2}}\\{ab - {c^2} - bc + {a^2}}&{bc - {a^2} - ca + {b^2}}&{ca - {b^2}}\end{array}} \right|$

and $\left. {{C_2} \to {C_2} - {C_3}} \right]$
$= \left| {\begin{array}{llllllllllllllllllll}{(b - a)(a + b + c)}&{(c - b)(a + b + c)}&{ab - {c^2}}\\{(c - b)(a + b + c)}&{(a - c)(a + b + c)}&{bc - {a^2}}\\{(a - c)(a + b + c)}&{(b - a)(a + b + c)}&{ca - {b^2}}\end{array}} \right|$

$= {(a + b + c)^2}\left| {\begin{array}{llllllllllllllllllll}{b - a}&{c - b}&{ab - {c^2}}\\{c - b}&{a - c}&{bc - {a^2}}\\{a - c}&{b - a}&{ca - {b^2}}\end{array}} \right|$

[taking $(a + b + c)$ common from ${C_1}$ and ${C_2}$

each]
$= {(a + b + c)^2}\left| {\begin{array}{cccccccccccccccccccc}0&0&{ab + bc + ca - \left( {{a^2} + {b^2} + {c^2}} \right)}\\{c - b}&{a - c}&{bc - {a^2}}\\{a - c}&{b - a}&{ca - {b^2}}\end{array}} \right|$

Now, expanding along ${R_1}$, $\left. { = {{(a + b + c)}^2}\left[ {ab + bc + ca - \left( {{a^2} + {b^2} + {c^2}} \right)} \right](c - b)(b - a) - {{(a - c)}^2}} \right]$

$= {(a + b + c)^2}\left( {ab + bc + ca - {a^2} - {b^2} - {c^2}} \right)$ $\left( {cb - ac - {b^2} + ab - {a^2} - {c^2} + 2ac} \right)$

$= {(a + b + c)^2}\left( {{a^2} + {b^2} + {c^2} - ab - bc - ca} \right)$

$\left( {{a^2} + {b^2} + {c^2} - ac - ab - bc} \right)$

$= \frac{1}{2}(a + b + c)\left[ {(a + b + c)\left( {{a^2} + {b^2} + {c^2} - ab - bc - ca} \right)} \right]$

$\left[ {{{(a - b)}^2} + {{(b - c)}^2} + {{(c - a)}^2}} \right]$

$= \frac{1}{2}(a + b + c)\left( {{a^3} + {b^3} + {c^3} - 3abc} \right)\left[ {{{(a - b)}^2} + {{(b - c)}^2} + {{(c - a)}^2}} \right]$

Hence, given determinant is divisible by $(a + b + c)$

and quotient is $\left( {{a^3} + {b^3} + {c^3} - 3abc} \right)\left[ {{{(a - b)}^2} + {{(b - c)}^2} + {{(c - a)}^2}} \right]$