Question

Using properties of determinants prove the following:
$\left|\begin{array}{ccc}1& 1& 1\\ a& b& c\\ a^3& b^3& c^3\end{array}\right|=(a-b)(b-c)(c-a)(a+b+c)$

Answer 1

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

$C_1\to C_1-C_2$

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

$C_2\to C_2-C_3$

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

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

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

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

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