<s>[INST] 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!
 [/INST] We have $$f(x)=ax^2+bx+c\\g(x)=bx^2+cx+a$$
If we set $t$ as the common root, then we know that $f(t)=g(t)=0$:
\begin{align}at^2+bt+c&=bt^2+ct+a\end{align}
We can equate coefficents to conclude that $a=b=c$
Therefore, we can say that \begin{align}\frac{a^3+b^3+c^3}{abc}&=\frac{a^3+a^3+a^3}{aaa}\\
&=\frac{3a^3}{a^3}\\
&=3\end{align}
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