Without expanding the determinant, prove that
$\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}a&{{a^2}}&{bc}\\b&{{b^2}}&{ca}\\c&{{c^2}}&{ab}\end{array}} \right| = \left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}1&{{a^2}}&{{a^3}}\\1&{{b^2}}&{{b^3}}\\1&{{c^2}}&{{c^3}}\end{array}} \right|$

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

Multiplying${R_1},{R_2}$ and ${R_3}$

by a, b and c respectively and dividing the determinant by abc,

we get
$\Delta = \cfrac{1}{{abc}}\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}{{a^2}}&{{a^3}}&{abc}\\{{b^2}}&{{b^3}}&{abc}\\{{c^2}}&{{c^3}}&{abc}\end{array}} \right|$

Taking abc common from ${C_3}$,

we get
$\Delta = \cfrac{{abc}}{{abc}}\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}{{a^2}}&{{a^3}}&1\\{{b^2}}&{{b^3}}&1\\{{c^2}}&{{c^3}}&1\end{array}} \right| = - \left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}{{a^2}}&1&{{a^3}}\\{{b^2}}&1&{{b^3}}\\{{c^2}}&1&{{c^3}}\end{array}} \right|$

Interchanging ${C_1} \leftrightarrow {C_2},$

we get $\Delta = \left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}1&{{a^2}}&{{a^3}}\\1&{{b^2}}&{{b^3}}\\1&{{c^2}}&{{c^3}}\end{array}} \right|$

Hence, $\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}a&{{a^2}}&{bc}\\b&{{b^2}}&{ca}\\c&{{c^2}}&{ab}\end{array}} \right| = \left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}1&{{a^2}}&{{a^3}}\\1&{{b^2}}&{{b^3}}\\1&{{c^2}}&{{c^3}}\end{array}} \right|$