Suppose $AB=BA$ and $A^{1965}=B^{2015}=I$. Prove that $A+B+I $ is invertible. Supppse $A $ abd $B $ are matrices, $AB=BA $ and $A^{1965}=B^{2015}=I $. Prove that $A+B+I $ is invertible.
I want to prove that $(A+B+I)C=I $ I have no idea how to start. Can any one give some hint?
 A: In case the hints in the comments were not enough, I have written up a complete solution.  Hover your mouse over to view it.  I wanted to break it up into multiple paragraphs, but I couldn't get the spoiler tags to work that way, so sorry if the formatting is a bit bad.

 Because there are no repeated roots in the equations of the form $x^n-1=0$, the minimal polynomials of $A$ and $B$ have no repeated roots, and hence $A$ and $B$ are both diagonalizable.  Since they commute, they are simultaneously diagonalizable, i.e., we can find a basis for our space of vectors that are both eigenvectors of $A$ and $B$.  Let $v$ be a common eigenvector of $A$ and $B$, so that $Av=av$ and $Bv=bv$.  We know that $a^{1965}=b^{2015}=1$.  Further, $v$ is an eigenvector of $A+B+I$ with eigenvalue $a+b+1$.  Since we have a basis of eigenvectors like this, showing that $A+B+I$ is invertible is equivalent to showing that $a+b+1\neq 0$.  Suppose that $a+b=-1$, where $a$ and $b$ lie on the unit circle in $\mathbb C$.  Since their imaginary parts cancel, $a$ and $b$ must be of the form $x+iy, z-iy$.  However, because there are at most $2$ points with a given $y$ coordinate on the unit circle, we must have that either $b=-a$ or $b=\overline{a}$.  In the first case, $a+b=0\neq -1$, and in the second case, $a,b=\frac{-1\pm\sqrt{-3}}{2}$, which are both primitive third roots of unity.

Now, as @Omnomnomnom remarked, $2015$ is not a multiple of $3$.  
