You want to divide both sides of the equation by some C such that \sin(\alpha)=\sqrt{3}/C and \cos(\alpha)=1/C. (So that we can use the trig summation identity)
Then it must be true that (\sqrt{3}/C)^2+(1/C)^2=1. It follows that C^2=3+1.
So in general, for y=A\cos(x)+B\sin(y), you want to...
This doesn't answer your question, but I find it to be another nice way of solving the problem.
Notice that \sqrt{3}^2+1^2=2^2.
So If you divide both sides of the equation by 2, you get...
The hint suggests to find a polynomial with integer coefficients. You have the right idea cubing, but there's a little more.
You found x_0^3=5+\sqrt[3]{6}(\sqrt[3]{2}+\sqrt[3]{3}).
But you can simplify that even more, by substituting x_0 in for \sqrt[3]{2}+\sqrt[3]{3}, to get...
This question amazes me. If we had said, 'no, it is unlikely that you will get the position,' would you not go? Bombed the interview on purpose? Cried?