I have proved a) and b) should I send you solutions?

You can post it right here, if you don't mind

$ \ ok \ !\ lets\ start!\ \\a) \ (x^2+y^2)^2\in Q \ so\ ,\ (x^4+y^4+2x^2y^2)\in Q \\ thus\ (xy)^2\in Q \\ next \ (x^6+y^6)=(x^2+y^2)(x^4+y^4-(xy)^2)\in Q \\ so\ (x^3+y^3)^2=(x^6+y^6+2(xy)^3)\in Q \\ which\ yiealds\ (xy)^3\in Q \\ finally\ \frac{(xy)^3}{(xy)^2}\in Q \ which\ means\ xy\in Q \ \\b) \ (x^3+y^3)=(x+y)(x^2+y^2-xy)\in Q \\ which\ means\ (x+y)\in Q $ solving c) but it seem not possible, I may be wrong))

For part c), I can assure you, that number $s_{1}$ can be irrational, if $s_{2}$ and $s_{3}$ are rational numbers

so i must find it)))