<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] $$0=ax^3+bx^2+cx=ax^3-a$$
which gives $x=1$, $a+b+c=0$ and since $$a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-ac-bc)=0,$$ we obtain:
$$\frac{a^3+b^3+c^3}{abc}=\frac{3abc}{abc}=3.$$
If you wish the solution for $x\in\mathbb C$ and $\{a,b,c\}\subset\mathbb C$ then since
$$a^3+b^3+c^3-3abc=(a+b+c)(a+\zeta b+\zeta^2c)(a+\zeta^2 b+\zeta c),$$
where $\zeta=-\frac{1}{2}+\frac{\sqrt{3}}{2}i$, we get the same answer.
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