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If $p\equiv 1$ mod $3$, I know that $x^{3}-6$ has any solution in $\mathbb{F}_{p}$ if and only if there exist $A,B\in\mathbb{Z}$ such that $p=A^{2}+3B^{2}$ and $9|B$ or $9|(2B+A)$ or $9|(2B-A)$ (This is one of the many conjectures of Euler). I have to show that, if this happens, then $x^{3}-6$ factors into different linear factors in $\mathbb{F}_{p}[x]$.

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The answer doesn’t have anything to do with $A$ and $B$. It’s just that the multiplicative group of $\mathbb F_p$ is cyclic of order divisible by $3$, thus has three cube roots of unity. If there’s one cube root of $6$, then there are three.

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