Question

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

Answer 1

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

$R_2\to R_2-R_3{R}_1\to R_1-R_2$

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

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

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

$R_2\to R_2-R_1$

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

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

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