find all x such that x^3-x+1 is congruent to 0 mod 6.
i tried but it seems like there is no x
A little observation should tell you that $\displaystyle x^3-x+1$ is always odd and therefore can never be divisible by $\displaystyle 6.$
If you delve further, you have $\displaystyle x^3-x=(x-1)x(x+1),$ a product of three consecutive integers – this is always divisible by $\displaystyle 6$ because it always contains exactly one multiple of $\displaystyle 3$ and at least one multiple of $\displaystyle 2.$ Hence $\displaystyle x^3-x+1\equiv1\mod6$ for all $\displaystyle x\in\mathbb Z.$