$\frac{a^{3}+1}{b+1}+\frac{b^{3}+1}{a+1}$ an integer $\Rightarrow \frac{a^{3}+1}{b+1}$ and $\frac{b^{3}+1}{a+1}$ are integers. I want to show that if the natural numbers $a,b \in \mathbb{N}$ are such that $\frac{a^3+1}{b+1}+\frac{b^3+1}{a+1} \in \mathbb{N}$, then, necessarily, $\frac{a^3+1}{b+1} \in \mathbb{N}$ and $\frac{b^3+1}{b+1} \in \mathbb{N}$.
I have thought the following.
We are given that $\frac{a^3+1}{b+1}+\frac{b^3+1}{a+1} \in \mathbb{N}$.
This means that $\frac{a^3+1}{b+1}+\frac{b^3+1}{a+1}=k, \text{ for some } k\in \mathbb{N}$.
We also have that $b+1 \mid a^3+1$ and $a+1 \mid b^3+1$, right?
So does it suffice to reject the cases that $a^3+1=k_1(b+1)$ and $b^3+1=k_2 (a+1)$ for negative $k_1, k_2$ ? If so, then we pick all the possible combinations and want to get a contradiction from the fact that $\frac{a^3+1}{b+1}+\frac{b^3+1}{a+1} \in \mathbb{N}$, or not?
Or do we show somehow else that $\frac{a^3+1}{b+1} \in \mathbb{N}$ and $\frac{b^3+1}{b+1} \in \mathbb{N}$ ?

Hint $\ r\! +\! s,\, rs \in \Bbb Z\,\Rightarrow\, r,s \in \Bbb Z\ $ by applying the Rational Root Test to $\,(x\!-\!r)(x\!-\!s)\in\Bbb Z[x]$
