Question

If lines ax+by+c=0,bx+cy+a=0 and cx+ay+b=0 are concurrent then

Answer 1

$ax+by+c=0$

$bx+cy+a=0$

$cx+ay+b=0$

Adding all the above equations, we get,

$ax+by+bx+cy+a+cx+ay+b=0$

$ax+ay+a+bx+by+b+cx+xy+c=0$

$a(x+y+1)+b(x+y+1)+c(x+y+1)=0$

$(x-y+1)(a+b+c)=0$

Assuming $(a+b+c)=0$, we get,

$a+b=-c.............\left(1\right)$

cubing both sides,

$\begin{array}{c}{a}^3+b^3+3ab\left(a+b\right)=-c^3\\ a^3+b^3+3ab\left(-c\right)=-c^3.........fromeq.\left(1\right)\\ a^3+b^3-3abc=-c^3\\ a^3+b^3+c^3=3abc\end{array}$