(a) $x^3-y^3\equiv 0\pmod{p}\implies x^2+xy+y^2\equiv 0\pmod{p}$. Since, $p>y-x>0$. Thus we get $\begin{cases}x^2+xy+y^2\equiv 0\pmod{p}\\ x^2+xz+z^2\equiv 0\pmod{p}\end{cases}$. Combining them we get $(y-z)(x+y+z)\equiv 0\pmod{p}\implies x+y+z\equiv 0\pmod{p}$ Note that using this we can prove the second statement. (b) Subtracting the first equation from the first we get. $(z-y)(z+y)\equiv x(y-z)\pmod{p}\implies x+y+z\equiv 0\pmod{p}$. Hence the result.

First part From the inequality we conclude $p|y^2+yx+x^2$ and $p|z^2+xz+x^2$ $\Rightarrow p|y^2-z^2+x(y-z)$ $\Rightarrow p|x+y+z$ (Using the inequality) $\Rightarrow p|(x+y+z)^2-(y^2+xy+x^2)-(z^2+xz+x^2)$ $\Rightarrow p|x^2-x(y+z)-2yz$ $\Rightarrow p|2(x^2-yz)$ (since $p|x+y+z$) Since there are three integers between $0$ and $p$, $p\neq 2$.Hence $p|x^2-yz$.Thus $p|(x+y+z)(x-z)-x^2+yz$ and $p|(x+y+z)(x-y)-x^2+yz$ $\Rightarrow p|z^2-xy$ and $p|y^2-xz$. Second part $p|y^2-xz$ and $p|z^2-xy$ $\Rightarrow p|y^2-z^2+x(y-z)$ $\Rightarrow p|(z-y)(x+y+z)$ $\Rightarrow p|x+y+z$ (Using the inequality) Also $p|y^2+z^2-x(y+z)$ $\Rightarrow p|x^2+y^2+z^2$ (Since $p|x+y+z$) $\Rightarrow p|x^2+y^2+z^2-(y^2-xz)$ and $p|x^2+y^2+z^2-(z^2-xy)$ $\Rightarrow p|x^2+xz+z^2$ and $p|x^2+xy+y^2$ $\Rightarrow p|x^3-z^3$ and $p|x^3-y^3$. Thus we get $p|x^3-y^3 $ and $p|x^3-z^3 \Leftrightarrow p|y^2-xz$ and $p|z^2-xy$.

From $(b)$ $ y^3\equiv xyz\pmod p$ $z^3\equiv xyz\pmod p$ Subtracting $y^3-z^3\equiv 0\pmod p$ Subtracting the given in $(a)$ $y^3-z^3\equiv 0\pmod p$ Thus the following statements are equivalent.