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Let $$ F(X)=X^6+aX^3+1 $$ Find all $a \in\Bbb C$ such that $F(X)$ has a multiple root.

I had this exercise in an exam, and I know I got the wrong answer but I don't know why my method didn't work.

What I think I should've done:

$$F(X)=X^6+aX^3+1$$ $$F'(X)=6X^5+3aX^2=3X^2(2X^3+a)$$

That means that $F'(c)=0 \equiv \left(c=0 \text{ or } c^3=\frac {-a} 2 \right)$

Plugging that into $F$ I get that $c=0$ doesn't work and with $c^3= \frac {-a} 2$ I get: $$F(c)= \frac {a^2}4 - \frac {a^2} 2+1=0 \iff a= \pm2$$

What I did:

$F$ has a multiple root then $\exists c : (x-c) \mid F \text{ and } (x-c)\mid F' \implies (x-c)\mid \gcd(F,F')$

That means that $\deg(\gcd(F,F'))\geq1$. I calculated the gcd and I got it was $1 \forall a\neq 0$, then I tried with $a=0$ and I also got $gcd=1$ which would mean there doesn't exist such a $c$.

Where did I go wrong here?

Also, how do I know that I got all the possible values for $a$ in the first one?

YoTengoUnLCD
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2 Answers2

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Since $F(X)=G(X^3)$, where $G(X)=X^2+aX+1$, if $\alpha$ is a root of $G$, then the cube roots of $\alpha$ are the roots of $F$.

If $G$ has multiple roots, then the same is of course true for $F$. Suppose $G$ has distinct roots $\alpha$ and $\beta$. If a cube root of $\alpha$ is the same as a cube root of $\beta$, then $\alpha=\beta$, a contradiction.

Check your computation of the greatest common divisor.

egreg
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It is clear that you made a mistake when computed the gcd and got that it was $1$ for all $a \ne 0$. This is so since as you already showed the gcd is not $1$ when $a = \pm 2$. You are sure that you got all the possible solutions since for any common root $c$ of $F$ and $F'$ you showed that $a=\pm 2$. Finally, if $a = \pm 2$ then $F(X) = (X^3 \pm 1)^2$ hence $F$ has multiple roots.

Holonomia
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