10. $[8]$ Find the largest positive integer $n$ such that $n^{3}+4 n^{2}-15 n-18$ is the cube of an integer.
Answer: 19 Note that the next cube after $n^{3}$ is $(n+1)^{3}=n^{3}+3 n^{2}+3 n+1$. After that, it is $(n+2)^{3}=n^{3}+6 n^{2}+12 n+8 \cdot n^{3}+6 n^{3}+12 n+8$ is definitely bigger than $n^{3}+4 n^{2}-15 n-18$, so the largest cube that $n^{3}+4 n^{2}-15 n-18$ could be is $(n+1)^{3}$. On the other hand, for $n \geq 4, n^{3}+4 n^{2}-15 n-18$ is larger than $(n-2)^{3}=n^{3}-6 n^{2}+12 n-8$ (as $4 n^{2}-15 n-18>-6 n^{2}+12 n-8 \Longleftrightarrow 10 n^{2}-27 n-10>0$, which is true for $n \geq 4$ ).
So, we will check for all solutions to $n^{3}+4 n^{2}-15 n-18=(n-1)^{3}, n^{3},(n+1)^{3}$. The first case yields
$$
n^{3}+4 n^{2}-15 n-18=n^{3}-3 n^{2}+3 n-1 \Longleftrightarrow 7 n^{2}-18 n-17=0
$$
which has no integer solutions. The second case yields
$$
n^{3}+4 n^{2}-15 n-18=n^{3} \Longleftrightarrow 4 n^{2}-15 n-18=0
$$
which also has no integer solutions. The final case yields
$$
n^{3}+4 n^{2}-15 n-18=n^{3}+3 n^{2}+3 n+1 \Longleftrightarrow n^{2}-18 n-19=0
$$
which has integer solutions $n=-1,19$. So, the largest possible $n$ is 19 .
Remark: The easiest way to see that the first two polynomials have no integer solutions is using the Rational Root Theorem, which states that the rational solutions of a polynomial $a x^{n}+\ldots+b$ are all of the form $\pm \frac{b^{\prime}}{a^{\prime}}$, where $b^{\prime}$ divides $b$ and $a^{\prime}$ divides $a$.