Question

Prove that: $\left|\begin{array}{ccc}1& 1& 1\\ a& b& c\\ a^3& b^3& c^3\end{array}\right|=(b-c)(c-a)(a-b)(a+b+c)$.

Answer 1

$\Delta =\left|\begin{array}{ccc}1& 1& 1\\ a& b& c\\ a^3& b^3& c^3\end{array}\right|C_2=C_2-C_1\Rightarrow \Delta =\left|\begin{array}{ccc}1& 0& 1\\ a& b-a& c\\ a^3& b^3-a^3& c^3\end{array}\right|C_3=C_3-C_1\Rightarrow \Delta =\left|\begin{array}{ccc}1& 0& 0\\ a& b-a& c-a\\ a^3& b^3-a^3& c^3-a^3\end{array}\right|$

Expanding along $R_1$

$\Rightarrow \Delta =1\left\{\right(b-a\left)\right(c^3-a^3)-(c-a\left)\right(b^3-a^3)-0+0\Rightarrow \Delta =(b-a\left)\right(c-a\left)\right(c^2+a^2-ac)-(c-a\left)\right(b-a\left)\right(b^2+a^2-ab)\Rightarrow \Delta =(b-a\left)\right(c-a\left)\right\{c^2+a^2-ac-b^2-a^2+ab\}\Rightarrow \Delta =(b-a\left)\right(c-a\left)\right\{c^2-b^2+a(b-c)\}\Rightarrow \Delta =(b-a\left)\right(c-a\left)\right\{(c-b)(c+b)+a(c-b)\}\Rightarrow \Delta =(b-a\left)\right(c-a\left)\right(c-b\left)\right(a+b+c)$