Verified answer

Answer:

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.............(1)

cubing both sides,

a3+b3+3ab(a+b)=−c3a3+b3+3ab(−c)=−c3.........fromeq.(1)a3+b3−3abc=−c3a3+b3+c