Let $R$ be a ring with unity where
$$x^3=x,\;\;\; \forall x \in R$$
How do I prove that $$x+x+x+x+x+x=0$$
Simply calculate $2 = 1+1 = (1+1)^3 = 8$ so $6 = 0$ (here I use an integer $n$ to mean the unit added to itself $n$ times). Now note that any element added to itself $6$ times is the same as $6$ times that element, which is then $0$.
Hint $\rm\,\ \forall x\!: f(x) = 0\:\Rightarrow\:\forall n\in \Bbb Z\!: f(n) = 0\ (in\ R)\:\Rightarrow\: char\, R\mid\, gcd(f(\Bbb Z))$
$(x + x)^3 = x^3 + 3x^3 + 3x^3 + x^3$ by the binomial theorem. Now use the condition that $x^3 = x$ for all elements in the ring to conclude that $2x = 8x$, from which the desired conclusion follows.