Without expanding the determinant, prove that

$\left|\begin{array}{ccc}a & a^{2} & b c \\ b & b^{2} & c a \\ c & c^{2} & a b\end{array}\right|=\left|\begin{array}{lll}1 & a^{2} & a^{3} \\ 1 & b^{2} & b^{3} \\ 1 & c^{2} & c^{3}\end{array}\right|$

L.H.S. $=\left|\begin{array}{lll}a & a^{2} & b c \\ b & b^{2} & c a \\ c & c^{2} & a b\end{array}\right|$

$=\frac{1}{a b c}\left|\begin{array}{ccc}a^{2} & a^{3} & a b c \\ b^{2} & b^{3} & a b c \\ c^{2} & c^{3} & a b c\end{array}\right|$     $\left[R_{1} \rightarrow a R_{1}, R_{2} \rightarrow b R_{2}\right.$, and $\left.R_{3} \rightarrow c R_{3}\right]$

$=\frac{1}{a b c} \cdot a b c\left|\begin{array}{lll}a^{2} & a^{3} & 1 \\ b^{2} & b^{3} & 1 \\ c^{2} & c^{3} & 1\end{array}\right| \quad$ [Taking out factor $a b c$ from $\mathrm{C}_{3}$ ]

$=\left|\begin{array}{lll}a^{2} & a^{3} & 1 \\ b^{2} & b^{3} & 1 \\ c^{2} & c^{3} & 1\end{array}\right|$

$=\left|\begin{array}{lll}1 & a^{2} & a^{3} \\ 1 & b^{2} & b^{3} \\ 1 & c^{2} & c^{3}\end{array}\right| \quad\left[\right.$ Applying $\mathrm{C}_{1} \leftrightarrow \mathrm{C}_{3}$ and $\mathrm{C}_{2} \leftrightarrow \mathrm{C}_{3}$ ]

$=$ R.H.S.

Hence, the given result is proved.