Question

Using the properties of determinants, show that:
(i)$\left|\begin{array}{ccc}1& a& a^2\\ 1& b& b^2\\ 1& c& c^2\end{array}\right|=(a-b)(b-c)(c-a)$
(ii)$\left|\begin{array}{ccc}1& 1& 1\\ a& b& c\\ a^3& b^3& c^3\end{array}\right|=(a-b)(b-c)(c-a)(a+b+c)$

Answer 1

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

Applying $R_2\to R_2-R_1$ and $R_3\to R_3-R_1$
$=\left|\begin{array}{ccc}1& a& a^2\\ 0& b-a& b^2-a^2\\ 0& c-a& c^2-a^2\end{array}\right|$

Taking $(b-a)$ common from $R_2$ and $c-a$ from $R_3$

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

Expanding along first column,
$=(b-a)(c-a)\left|\begin{array}{cc}1& b+a\\ 1& c+a\end{array}\right|$
$=(b-a)(c-a)(c+a-b-a)$
$=(b-a)(c-a)(c-b)$

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

$=RHS$

(ii) Consider, $LHS=\left|\begin{array}{ccc}1& 1& 1\\ a& b& c\\ a^3& b^3& c^3\end{array}\right|$

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

$=\left|\begin{array}{ccc}0& 0& 1\\ a-b& b-c& c\\ a^3-b^3& b^3-c^3& c^3\end{array}\right|$

Taking $(a-b)$ as a common factor from $C_1$ and $b-c$ as common factor from $C_2$

$=(a-b)(b-c)\left|\begin{array}{ccc}0& 0& 1\\ 1& 1& c\\ a^2+ab+b^2& b^2+bc+c^2& c^3\end{array}\right|$

Expanding along $R_1$, we get

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

$=(a-b)(b-c)(b^2+bc+c^2-a^2-ab-b^2)$

$=(a-b)(b-c)[bc-ab+(c^2-a^2\left)\right]$

$=(a-b)(b-c[b(c-a)+(c-a)(c+a)]$

$=(a-b)(b-c)c-a\left)\right(b+c+a)$