If $A^3 = I_n$ then $\operatorname{tr}(A)\in\Bbb Z$

Hint: use the fact that all irreducible factors of the characteristic polynomial of $A$ divide its minimal polynomial. (See Minimal and characteristic polynomial have same set of irreducible factors)

The minimal polynomial of $A$ divides $x^3-1$, so the irreducible factors of the minimal polynomial have all integer coefficients. (The only possible factors are $x-1$ and $x^2+x+1$.) Hence the characteristic polynomial of $A$ is a product of polynomials with integer coefficients. In particular, $\operatorname{Tr}(A) \in \mathbb Z$.

Hint If $\lambda$ is an eigenvalue of $A$ then $\lambda \in \{1, \omega, \overline{\omega} \}$ where $\omega^2+\omega+1=0$.

Let $m,n,k$ be the multiplicities of these eigenvalues (which could be 0).

Then $$\mbox{tr}(A)=m+n \omega+k \bar{\omega}$$

Use the fact that $A$ has real entries (or that $\mbox{tr}(A)$ is real) to show that $m=k$. Then use the fact that $\omega+\bar{\omega}=-1$.