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By using properties of determinants, show that:
$\left|\begin{array}{lll}1 & a & a^{2} \\ 1 & b & b^{2} \\ 1 & c & c^{2}\end{array}\right|=(a+b)(b-c)(c-a)$
Solution
Let $\Delta=\left|\begin{array}{lll}1 & a & a^{2} \\ 1 & b & b^{2} \\ 1 & c & c^{2}\end{array}\right|$
Applying $R_{1} \rightarrow R_{1}-R_{3}$ and $R_{2} \rightarrow R_{2}-R_{3},$ we have:
$\Delta=\left|\begin{array}{ccc}0 & a-c & a^{2}-c^{2} \\ 0 & b-c & b^{2}-c^{2} \\ 1 & c & c^{2}\end{array}\right|$
$=(c-a)(b-c)\left|\begin{array}{ccc}0 & -1 & -a-c \\ 0 & 1 & b+c \\ 1 & c & c^{2}\end{array}\right|$
Applying $R_{1} \rightarrow R_{1}+R_{2},$ we have:
$\Delta=(b-c)(c-a)\left|\begin{array}{ccc}0 & 0 & -a+b \\ 0 & 1 & b+c \\ 1 & c & c^{2}\end{array}\right|$
$=(a-b)(b-c)(c-a)\left|\begin{array}{ccc}0 & 0 & -1 \\ 0 & 1 & b+c \\ 1 & c & c^{2}\end{array}\right|$
Expanding along $\mathrm{C}_{1},$ we have:
$\Delta=(a-b)(b-c)(c-a)\left|\begin{array}{cc}0 & -1 \\ 1 & b+c\end{array}\right|=(a-b)(b-c)(c-a)$
Hence, the given result is proved.
