### Instruction: Let
\[\mathbf{M} = \begin{pmatrix} a & b \\ c & d \end{pmatrix}\]be a matrix with real entries such that $\mathbf{M}^3 = \mathbf{I}.$  Enter all possible values of $a + d,$ separated by commas.### Response: The problem asks me to find the trace of $\mathbf{M},$ which is the sum of its diagonal entries.
I know that the trace of a product of matrices is equal to the product of their traces, so I can write \[\text{tr}(\mathbf{M}^3) = \text{tr}(\mathbf{M})^3.\]
Since $\mathbf{M}^3 = \mathbf{I},$ I know that $\text{tr}(\mathbf{M}^3) = \text{tr}(\mathbf{I}) = 2.$
Therefore, I have \[\text{tr}(\mathbf{M})^3 = 2.\]
Taking the cube root of both sides, I get \[\text{tr}(\mathbf{M}) = \sqrt[3]{2}.\]
This means that $a + d = \sqrt[3]{2}.$
However, this is not the only possible value of $a + d.$
I notice that $\mathbf{M}$ could also be a rotation matrix of the form \[\mathbf{M} = \begin{pmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{pmatrix},\] where $\theta$ is some angle.
In that case, $\mathbf{M}^3$ would be a rotation by $3\theta,$ and if $3\theta$ is a multiple of $2\pi,$ then $\mathbf{M}^3 = \mathbf{I}.$
So, another possible value of $a + d$ is $2\cos \theta,$ where $\theta$ is a multiple of $\frac{2\pi}{3}.$
The multiples of $\frac{2\pi}{3}$ are $\frac{2\pi}{3},$ $\frac{4\pi}{3},$ $\frac{6\pi}{3},$ etc.
The corresponding values of $2\cos \theta$ are $-1,$ $-1,$ $2,$ etc.
Therefore, the possible values of $a + d$ are $\sqrt[3]{2},$ $-1,$ and $2.$