Question

Prove that the lines
$ax+by+c=0,bx+cy+a=0$
and $cx+ay+b=0$ are concurrent
$a^3+b^3+c^3=3abc$
or if $a+b+c=0$.

Answer 1

The line will be concurrent if
$\left|\begin{array}{ccc}a& b& c\\ b& c& a\\ c& a& b\end{array}\right|=0$
Or $a(bc-a^2)-b(b^2-ac)+c(ab-c^2)=0$
Or $a^3{+b}^3{+c}^3-3abc=0$
Or $(a+b+c)(a^2+b^2+c^2-ab-bc-ca)=0$
Or $a+b+c=0$
as the other factor is always $+$ive $=\frac{1}{2}\sum {(a-b)}^2$