If $a^4+b^4\in\mathbb Q$ and $a^3+b^3\in\mathbb Q$ and $a^2+b^2\in\mathbb Q$, prove that $a+b\in\mathbb Q$ and $ab\in\mathbb Q$. 
If $\begin{cases}a^4+b^4\in\mathbb Q\\ a^3+b^3\in\mathbb Q\\ a^2+b^2\in\mathbb Q\end{cases}$, prove that $a+b\in\mathbb Q$ and $ab\in\mathbb Q$. It is given that $a,b\in\mathbb R$.

The proof of the latter would simply follow from the former, and vice versa. So I think a better question would be:

Prove one of these statements: $a+b\in\mathbb Q$ or $ab\in\mathbb Q$.

The problem is from the selection to IMO.
I've tried a whole lot of things, including the identities: $$a^4+b^4=(a+b)(a^3+b^3)-ab(a^2+b^2)\\ a^3+b^3=(a+b)(a^2-ab+b^2)\\ a^2+b^2=(a+b)^2-2ab\\ (a+b)^3=a^3+b^3+3ab(a+b)\\ \text{etc...}$$
Even if one could solve the problem using these identities, doing it would most likely be quite tedious imho... Any observations would be greatly appreciated. Thanks.

(Note, added later:  This answer was posted before the OP added the stipulation that $a,b\in\mathbb{R}$.)
This is more plodding than ronno's slick hint, but perhaps it shows more of the thought process:
$$(a^2+b^2)^2=a^4+2(ab)^2+b^4\implies (ab)^2\in\mathbb{Q}$$
$$(a^2+b^2)^3=a^6+3(ab)^2(a^2+b^2)+b^6\implies a^6+b^6\in\mathbb{Q}$$
$$(a^3+b^3)^2=a^6+2(ab)^3+b^6\implies (ab)^3\in\mathbb{Q}$$
$$(ab)^2\in\mathbb{Q}\land (ab)^3\in\mathbb{Q}\implies ab\in\mathbb{Q}$$
It would be nice to prove $a+b\in\mathbb{Q}$ as well, but you can't:  $a=1+\sqrt{-3}, b=-1+\sqrt{-3}$ is a counterexample.  What you can show is
$$a^3+b^3=(a+b)(a^2-ab+b^2)\implies a+b\in\mathbb{Q}\lor a^2-ab+b^2=0$$
which is to say, $a+b$ is rational unless $a^3=-b^3$ with $a\not=-b$.
