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Here is the problem:

$P(x)$ and $Q(x)$ are polynomials with real coefficients and $P(x)^3-Q(x)^3$ is divisible by $(x-a)^2$ but not by $(x-a)^3$. Prove that $P(x)-Q(x)$ is divisible by $(x-a)^2$.

I am pretty sure this problem is not that hard when you figure out how to go about solving it, but I'm having trouble with it. I am pretty sure that it requires some kind of application of the fundamental theorem of algebra, as $(x-a)$ is obviously supposed to be a root of the polynomial made by $P(x)-Q(x)$. Yet I am not sure how to go about proving the property in question. Any help would be appreciated. Thank you in advance.

suomynona
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1 Answers1

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$$P^3-Q^3 = (P-Q)(P^2+PQ+Q^2)$$

If $a$ was a root of $P^2+PQ+Q^2$ then

$$P(a)^2+Q(a)^2 = -P(a)Q(a)$$

So $P(a)Q(a) \le 0$. On the other hand,

$$\left[P(a)+Q(a)\right]^2 = P(a)^2+2P(a)Q(a)+Q(a)^2 = P(a)Q(a)$$

So that $P(a)Q(a)\ge 0$. So $P(a)Q(a)=0$ and $a$ must be a root of either $P$ or $Q$. Say $a$ was a root of $P$, then $P^3(a)=0$ and $P^3(a)-Q^3(a)=0$ implies $Q(a)=0$, hence $a$ is also a root of $Q$.

This implies $(x-a)^3$ divides $P^3(x)-Q^3(x)$, since $a$ is a root of multiplicity at least 1 for $P-Q$ and of at least multiplicity 2 for $P^2+PQ+Q^2$.

Therefore $a$ is not a root of the factor $P^2+PQ+Q^2$, so both factors of $(x-a)$ must belong to $P-Q$

David P
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