Find all monic polynomials $P, Q$ in $\mathbb{Q}[X]$ such that:

$$
P(x)^{3}+Q(x)^{3}=x^{12}+1
$$
We start by factoring, which gives:

$$
(P(x)+Q(x))\left(P^{2}(x)-P(x) Q(x)+Q^{2}(x)\right)=\left(x^{8}-x^{4}+1\right)\left(x^{4}+1\right)
$$

We recognize cyclotomic polynomials, which are irreducible. Since $P, Q$ are monic, the degree of $P+Q$ is greater than 1, and the degree of $P^{2}-P Q+Q^{2}$ is at least $2 \cdot \max (\operatorname{deg}(P), \operatorname{deg}(Q))$, so the degree of the second factor is greater than the first. Moreover, we recognize two cyclotomic polynomials, which are therefore irreducible in $\mathbb{Q}[x]$. Thus, the only possibility to satisfy the degree conditions is:

$$
\left\{\begin{array}{l}
P+Q=\frac{1}{C}\left(x^{4}+1\right) \\
P^{2}-P Q+Q^{2}=C\left(x^{8}-x^{4}+1\right)
\end{array}\right.
$$

The leading coefficient of $P^{2}-P Q+Q^{2}$ is always 1, so $C=1$. We then have:

$$
3 P Q=(P+Q)^{2}-\left(P^{2}-P Q+Q^{2}\right)=\left(x^{4}+1\right)^{2}-\left(x^{8}-x^{4}+1\right)=3 x^{4}
$$

If $x \mid P, Q$, then $x \mid x^{12}+1$, which is absurd. Therefore, $P=1$ and $Q=x^{4}$ or vice versa.

Key ideas: Always think about factoring a polynomial when possible. Look at the highest and lowest degree coefficients.