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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 $$a^3+b^3+c^3$$

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Are you sure your second equation is $c^2+bx+a = 0$? – Emily Aug 9 '12 at 1:23
@EdGorcenski Damn I am sorry – HackToHell Aug 9 '12 at 1:29
Hey @Mark, dunno if you know this, but a number of us have actually been trying to kill the algebra tag for a while now... – J. M. Aug 9 '12 at 2:25
@J.M. Noted, thanks. – MJD Aug 9 '12 at 2:33
up vote 8 down vote accepted

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.

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It is the answer indeed, thanks – HackToHell Aug 9 '12 at 12:44

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



Cubing both sides, $(a+b)^3=(-c)^3$


or, $=>a^3+b^3+3ab(-c)=-c^3$

$=>a^3+b+c^3=3abc$ if a≠c.

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