A is an n by n matrix such that $A^3=2I$. A is an n by n matrix such that $A^3 = 2I$.
(1) Is $A$ invertible? If it is, then what is $A^{-1}$?
(2) Is $A+I$ invertible? If it is, then what is $(I+A)^{-1}$
The original version of the question is not in English, so maybe there are some grammar mistakes, sorry in advance.

A matrix $B$ that satisfies $AB=BA = I$ is an inverse of $A$.
Since $A ({1 \over 2} A^2) = I$ we see that $A^{-1} = {1 \over 2} A^2$.
Suppose $x^3 = 2$, then we see that
${1 \over x+1} = {1 \over x} (1-{1 \over x} + {1 \over x^2} - {1 \over x^3} + {1 \over x^4} - {1 \over x^5} +\cdots) $, and using $x^3 = 2$ we get
$=1+(-1+{1 \over x} - {1 \over x^2}) (1-{1 \over 2} + {1 \over 2^2} - \cdots)$ and using ${x^3 \over 2} =1$ we get
$= {1 \over 3} (1-x+x^2)$.
(Check by multiplying by $1+x$.)
This suggests trying $B={1 \over 3} (I-A+A^2)$ as an inverse for $I+A$.
As above, if we multiply $(I+A)B$ we get $I$ hence
$(I+A)^{-1} = {1 \over 3} (I-A+A^2)$.
Addendum: Another approach to computing $(A+I)^{-1}$.
We want to find a polynomial $p$ such that $(x+1) p(x)$ is a non zero constant. Since $x^3=2$ we posit that a degree two polynomial will suffice. Hence we are looking for $a,b$ such that $(x+1) (1 + ax+bx^2)$ is a constant.
Expanding (and using $x^3=2$) gives
$a=-1, b=1$ which is the same as above.
