Find all rational numbers $x$ such that $x^{3}+(x+1)^{3}+(x+2)^{3}=(x+3)^{3}$.
After simplification, it is a matter of finding the rational roots of the polynomial $P(x)=x^{3}-6 x-9$. We notice that 3 is a root of $P$. By polynomial division, $P(x)=(x-3)\left(x^{2}+3 x+3\right)$. Since $x^{2}+3 x+3$ has no real solutions $(\Delta=-3), P$ has no other rational roots than 3.

Another idea is to exploit the following lemma:

Lemma 95. Let $P(X)=a_{n} X^{n}+\cdots+a_{1} X+a_{0}$ be a polynomial with integer coefficients for which we are looking for rational roots in the form $x=\frac{r}{s}$, with $r$ and $s$ coprime. Then:

$$
s \mid a_{n} \text { and } r \mid a_{0}
$$

## Proof.

$P(x)=0 \Leftrightarrow a_{n} \frac{r^{n}}{s^{n}}+a_{n-1} \frac{r^{n-1}}{s^{n-1}}+\cdots+a_{1} \frac{r}{s}+a_{0}=0 \Leftrightarrow a_{n} r^{n}+a_{n-1} r^{n-1} s+\cdots+a_{1} r s^{n-1}+a_{0} s^{n}=0$ Since $s \mid 0$, we have $s \mid a_{n} r^{n}+a_{n-1} r^{n-1} s+\cdots+a_{1} r s^{n-1}+a_{0} s^{n}$. Thus, $s \mid a_{n} r^{n}$. Since $s$ and $r$ are coprime, it follows that $s \mid a_{n}$. We similarly show that $r \mid a_{0}$.

According to our lemma, if $x=\frac{r}{s}$ is a rational root of $x^{3}-6 x-9$, then $s$ divides the leading coefficient of this polynomial, that is, $s \mid 1$ and thus $s=1$. In other words, all solutions to the problem are integers. And we even know that $r \mid 9$, hence $x \in\{-9,-3,-1,1,3,9\}$. We then easily verify that $x=3$ is the only value that works.