If $ax^2+bx+c = 0$ and $bx^2+cx + a = 0$ have a common root and $a\neq 0$, then find $\frac{a^3+b^3+c^3}{abc}$ If $ax^2+bx+c = 0$ and $bx^2+cx + a = 0$ have a common root and $a\neq 0$, then find $$\frac{a^3+b^3+c^3}{abc}$$
I tried that for both equations to have a common root, the expression on left hand sides must be equal, ie $$ax^2+bx+c = bx^2+cx + a$$
for this we must have $x=1$ (i cannot prove this, but it appears to be true). Also both of these must be equal to $0$, so we have:
$$a+b+c=0$$
So using this we say 
$$\frac{a^3+b^3+c^3}{abc} = \frac{a^3+b^3+c^3-3abc+3abc}{abc}=\frac{(a+b+c)(...)}{abc}+3$$
So we get the answer as $3$. 
How do we say that $x =1$ is the commmon root? Thanks!

HINT:
Let $t$ be the common root
So, we have $$at^2+bt+c=0\  \ \  \ (1)$$ and $$bt^2+ct+a=0\  \ \  \ (2)$$
So, we have two simultaneous equations in $t^2,t$
Solve for $t,t^2$ and use the identity $$t^2=(t)^2$$
