How do I proceed with these quadratic equations? The question is 
$$ax^2 + bx + c=0 $$ and $$cx^2+bx+a=0$$ have a common root, if $b≠ a+c$, then what is $$a^3+b^3+c^3$$ 
 A: The value of $a^3+b^3+c^3$ is not determined. Just choose $a=c$. But leaving out the condition $a\ne c$ is probably an oversight, so assume from now on that $a\ne c$. 
If $q$ is a common root, then $aq^2+bq+c=cq^2+bq+a=0$. Subtracting, we find that $(a-c)q^2-(a-c)=0$. Since $a\ne c$, we get $q^2=1$.
We cannot have $q=-1$, for that implies that $b=a+c$. We are left with $q=1$, which gives $a+b+c=0$. Conversely, if $a+b+c=0$, then $1$ is a common root of the two equations. 
That still leaves many possibilities for the value of $a^3+b^3+c^3$. For example, let $a=1$, $b=-3$, $c=2$. Then $a^3+b^3+c^3=-18$. Let $a=1$, $b=-4$, $c=2$. Then $a^3+b^3+c^3=-36$.  
However, we can say something interesting about $a^3+b^3+c^3$ in the case $a\ne c$.
Use the general identity
$$a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-bc-ca-ab),\tag{$1$}$$
which can be verified by multiplying out.   Since in our case $a+b+c=0$, we conclude that
$$a^3+b^3+c^3=3abc.$$
Perhaps this is the intended answer. 
A: If a=c, the equations become identical, we can hardly determine any relationship among a,b,c(=a).
If a≠c, 
if y is a root of $ax^2+bx+c=0$, then observe that $\frac{1}{y}$ is a root of $cx^2+bx+a=0$.
For the common root, $y=\frac{1}{y}=>y=±1$, but y=-1 makes a-b+c=0 ⇔ b=a+c which is not acceptable according to the given condition (as André Nicolas has observed).
So, y=1 if a≠c
=>a+b+c=0
=>a+b=-c
Cubing both sides, $(a+b)^3=(-c)^3$
$=>a^3+b^3+3ab(a+b)=-c^3$
or, $=>a^3+b^3+3ab(-c)=-c^3$
$=>a^3+b+c^3=3abc$ if a≠c.
