Part a) is trivial: $x^2 = \frac{xy \cdot zx}{yz}$ Then $x^2,y^2,z^2$ are rational numbers. Let $k = x^3+y^3+z^3 \in \mathbb{Q}$. Then $(k-x^3)^2 = (y^3+z^3)^2 = (y^2)^3 + (z^2)^3 + 2(yz)^3 \in \mathbb{Q}$ $k^2 + x^6 -2kx^3 \in \mathbb{Q}$, but $k, x^2 \in \mathbb{Q}$. Therefore $2kx^3 \in \mathbb{Q} \Rightarrow x^3 \in \mathbb{Q} \Rightarrow x = \frac{x^3}{x^2} \in \mathbb{Q}$. In the same way we prove that $y,z \in \mathbb{Q}$.

Hi, Edriv what about the case where k=0? in fact this cannot occur because of Fermat's last Theorem (n=3) and the fact that x/z and y/z are both rational numbers.

$\frac{x}{y}$ (and same about other pairs) is rational, so as we can represent an irrartional number by $Aa$ such that $A$ is rational and $a$ is irrational,which it cannot devide into productions of an irrational and a rational, so let $x=Aa,y=Bb,Z=Cc;$ by what we have at the first we will get that $a=b=c$,the rest is obviouse.

From part a) it follows that the numbers $x^2, y^2, z^2,\frac{x}{ y}, \frac{y}{ z}, \frac{z}{ x} $ are rational, and since $\frac{x^3+y^3+z^3}{y}=x^2(\frac{x}{ y})+y^2+z^2(\frac{z}{y})$, $y$ is rational and so are also $x, z$.