How to solve $x^3 \equiv 1 \pmod{37}$ We are asked to solve $x^3 \equiv 1 \pmod{37}$. I know that the answer is $10$ since $27\cdot37 = 999$ and $10^3 = 1000$ but how do I show this rigorously? If it helps, we are given the primitive roots of $37$ which are $2, 5, 13, 15, 17, 18, 19, 20, 22, 24, 32$, and $35$. But I am not sure how this is useful.
You know that $2^{36}\equiv 1$, since $2$ is a primitive root.  Hence you may conclude that $(2^{12})^3\equiv 1$, and also $(2^{24})^3\equiv (2^{36})^2\equiv 1^2$.  These correspond to $26$ and $10$ respectively.  Of course $2^{36}=1$.

Now, suppose that $(2^a)^3\equiv 1$.  Then $2^{3a}\equiv 1$, so $36|3a$, or $12|a$.  That doesn't leave many choices for $a$, since $1\le a\le 36$...
Just for fun another way.

Because of $\mathbb F_{37}$ is a field the equation in $\mathbb F_{37}[x]$ $$x^3=1\iff (x-1)(x^2+x+1)=0$$

must have three roots so, since $1$ is clearly root, one can find out the other two ones solving $$x^2+x+1=0$$ We have
$$x+y=-1=36\\xy=1$$ Exploring $$35+1\\34+2\\33+3\\32+4\\31+5\\30+6\\29+7\\28+8\\27+9\\26+10$$ We stop because $26\cdot10=260\equiv 1\pmod{37}$, thus the roots are $1,10$ and $26$.