By using properties of determinants,show that:
$\left|\begin{array}{lll}1 & a & a^{2} \\ 1 & b & b^{2} \\ 1 & c & c^{2}\end{array}\right|=(a-b)(b-c)(c-a)$

(N/A) Let $\Delta = \left|\begin{array}{lll}1 & a & a^{2} \\ 1 & b & b^{2} \\ 1 & c & c^{2}\end{array}\right|$.
Applying the row operations $R_{1} \rightarrow R_{1}-R_{2}$ and $R_{2} \rightarrow R_{2}-R_{3}$:
$\Delta = \left|\begin{array}{ccc}1-1 & a-b & a^{2}-b^{2} \\ 1-1 & b-c & b^{2}-c^{2} \\ 1 & c & c^{2}\end{array}\right| = \left|\begin{array}{ccc}0 & a-b & (a-b)(a+b) \\ 0 & b-c & (b-c)(b+c) \\ 1 & c & c^{2}\end{array}\right|$.
Taking out common factors $(a-b)$ from $R_{1}$ and $(b-c)$ from $R_{2}$:
$\Delta = (a-b)(b-c) \left|\begin{array}{ccc}0 & 1 & a+b \\ 0 & 1 & b+c \\ 1 & c & c^{2}\end{array}\right|$.
Applying $R_{2} \rightarrow R_{2}-R_{1}$:
$\Delta = (a-b)(b-c) \left|\begin{array}{ccc}0 & 1 & a+b \\ 0 & 0 & c-a \\ 1 & c & c^{2}\end{array}\right|$.
Expanding along $C_{1}$:
$\Delta = (a-b)(b-c) \cdot 1 \cdot \left|\begin{array}{cc}1 & a+b \\ 0 & c-a\end{array}\right| = (a-b)(b-c)(c-a)(1-0) = (a-b)(b-c)(c-a)$.
Hence,the result is proved.