If $x^3 =x$ then $6x=0$ in a ring

Let $R$ be a ring with unity where

$$x^3=x,\;\;\; \forall x \in R$$

How do I prove that $$x+x+x+x+x+x=0$$

Solution 1:

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 $0$.

Solution 2:

Hint $\rm\,\ \forall x\!: f(x) = 0\:\Rightarrow\:\forall n\in \Bbb Z\!: f(n) = 0\ (in\ R)\:\Rightarrow\: char\, R\mid\, gcd(f(\Bbb Z))$

Solution 3:

$(x + x)^3 = x^3 + 3x^3 + 3x^3 + x^3$ 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.