<s>[INST] Suppose that $P(x)$ is a polynomial of degree $n$ such that $P(k)=\frac{k^{2}}{k^{3}+1}$ for $k=0,1,\ldots,n$, find the value of $P(n+1)$ Suppose that $P(x)$ is a polynomial of degree $n$ such that $P(k)=\dfrac{k^{2}}{k^{3}+1}$ for $k=0,1,\ldots,n$. Find the value of $P(n+1)$
I tried by making a $f(x) = (x^3+1) P(x) - x^2$. But this equation will have $n+3$ roots, out of which $n+1$ will be $0,1,2,...,n$. What about the other two roots? Can we simplify $x^2 \over x^3+1$ more to help?
 [/INST] If $p(x)$ is a polynomial with degree $\leq n$ and $\delta$ is the forward difference operator, $(\delta p)(x)=p(x+1)-p(x)$, we have that
$$ \left(\delta^{n+1}p\right)(x) = \sum_{k=0}^{n+1}\binom{n+1}{k}(-1)^k\,p(x+k) = 0\tag{1} $$
hence in our case 
$$ P(n+1)=(-1)^n\sum_{k=0}^{n}\binom{n+1}{k}(-1)^k \frac{k^2}{k^3+1}.\tag{2} $$

Addendum. Can we compute a closed form for the RHS of (2)? With a bit of patience, yes.
Let $\omega=\exp\left(\frac{\pi i}{3}\right)=\frac{1+i\sqrt{3}}{2}$. By partial fraction decomposition we have
$$ \frac{k^2}{k+1} = \frac{1}{3}\left(\frac{1}{k+1}+\frac{1}{k-\omega}+\frac{1}{k-\overline{\omega}}\right)\tag{3} $$
and since $\frac{1}{k+1}=\int_{0}^{1}x^k\,dx$,
$$ \sum_{k=0}^{n}\binom{n+1}{k}\frac{(-1)^k}{k+1} = \frac{1+(-1)^n}{n+2}\tag{4} $$
$$ \sum_{k=0}^{n}\binom{n+1}{k}(-1)^k\left(\frac{1}{k-\omega}+\frac{1}{k-\overline{\omega}}\right)\\ = \frac{(-1)^n(2n+1)}{n^2+n+1}+\underbrace{2\,\text{Re}\,B\left(n+2,\frac{-1+i\sqrt{3}}{2}\right)}_{\text{wildly oscillating term.}}\tag{5} $$
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