Maximum value of $n$ for which $8n^3+16n^2+72n+64$ is a perfect cube

Question

What is the maximum value of $n$ for which $$8n^3+16n^2+72n+64$$ is a perfect cube?

I know how this could be done for a quadratic but how to extend it for a cubic.

Answer 1

Hint 1: $8n^3+16n^2+72n+64 = 2^3(n^3+2n^2+9n+8).$

Hint 2: $n^3+2n^2+9n+8>n^3$, so it can't be perfect cube if $(n+1)^3>n^3+2n^2+9n+8.$

Answer 2

I shall find all integers (not necessarily positive) $n$ such that $$8n^3+16n^2+72n+64=2^3\,\left(n^3+2n^2+9n+8\right)$$ is the cube of an integer. Clearly, $p(n):=n^3+2n^2+9n+8$ must be a perfect cube. Note that $$p(n)-(n-1)^3=5n^2+6n+9=\frac{(5n+3)^2+36}{5}>0$$ and $$(n+3)^3-p(n)=7n^2+18n+19=\frac{(7n+9)^2+52}{7}>0\,.$$ Thus, $$(n-1)^3<p(n)<(n+3)^3\,.$$

Therefore, $p(n)=n^3$, $p(n)=(n+1)^3$, or $p(n)=(n+2)^3$. In the first case, $p(n)=n^3$ iff $2n^2+9n+8=0$, or equivalently, $n=\dfrac{-9\pm\sqrt{17}}{4}$, none of whose values is an integer. In the second case, $p(n)=(n+1)^3$ iff $n^2-6n-7=0$, or equivalently, $n=-1$ or $n=7$. In the final case, $p(n)=(n+2)^3$ iff $4n^2+3n=0$, which is the same as saying that $n=0$ or $n=-\dfrac{3}{4}$. Consequently, all integer solutions are $n=-1$, $n=0$, and $n=7$.