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

(A) $L.H.S. = \left|\begin{array}{lll}a & a^{2} & b c \\ b & b^{2} & c a \\ c & c^{2} & a b\end{array}\right|$
Multiply $R_{1}$ by $a$,$R_{2}$ by $b$,and $R_{3}$ by $c$:
$= \frac{1}{abc} \left|\begin{array}{lll}a^{2} & a^{3} & abc \\ b^{2} & b^{3} & abc \\ c^{2} & c^{3} & abc\end{array}\right|$
Taking out the common factor $abc$ from $C_{3}$:
$= \frac{abc}{abc} \left|\begin{array}{lll}a^{2} & a^{3} & 1 \\ b^{2} & b^{3} & 1 \\ c^{2} & c^{3} & 1\end{array}\right| = \left|\begin{array}{lll}a^{2} & a^{3} & 1 \\ b^{2} & b^{3} & 1 \\ c^{2} & c^{3} & 1\end{array}\right|$
Apply $C_{1} \leftrightarrow C_{3}$ and then $C_{2} \leftrightarrow C_{3}$:
$= \left|\begin{array}{lll}1 & a^{2} & a^{3} \\ 1 & b^{2} & b^{3} \\ 1 & c^{2} & c^{3}\end{array}\right| = R.H.S.$
Hence,the result is proved.