Question

Show that $\left|\begin{array}{ccc}1& 1& 1\\ a& b& c\\ a^2& b^2& c^2\end{array}\right|=(a-b)(b-c)(c-a)$.

Answer 1

$\left|\begin{array}{ccc}1& 1& 1\\ a& b& c\\ a^2& b^2& c^2\end{array}\right|=(a-b)(b-c)(c-a)$

Solving determinant, we have

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

Applying $C_2\to C_2-C_1$ and $C_3\to C_3-C_1$

$=\left|\begin{array}{ccc}1& 0& 0\\ a& b-a& c-a\\ a^2& b^2-a^2& c^2-a^2\end{array}\right|$

Taking $(c-a)$ and $(b-a)$ as common from $C_3$ and $C_2$ respectively.

$=(b-a)(c-a)\left|\begin{array}{ccc}1& 0& 0\\ a& 1& 1\\ a^2& b+a& c+a\end{array}\right|$

Expanding the determinant along $R_1$, we have

$=(b-a)(c-a)(1(c+a-b-a)-0+0)$

$=(b-a)(c-a)(c-b)$

$=(a-b)(b-c)(c-a)$

$=$ R.H.S.

Hence proved.A