Only Cubes

Let $f(n)=n^3+2n^2+9n+8$ .

First note that $f(n)>n^3$ for all positive integer $n$ .

Now consider $(n+1)^3-f(n)=n^2-6n-7=(n+1)(n-7)$ . This has roots $n=-1$ and $n=7$ ; for all $n>7$ , it is positive, so for $n>7$ , we have $n^3<f(n)<(n+1)^3$ , and $f(n)$ can't be a cube.

It's easy to check that among $n=1,\ldots,7$ , the only cube is $f(7)=512=8^3$ . This proves that $n=\boxed7$ is the only positive integer solution.

$n^3+2n^2+9n+8=(n+1)(n^2+n+8)$

Assume $(n+1)^2=(n^2+n+8)$ , then $n=7$

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This does not prove that $n=7$ is the only solution.