Show that $A$ is an invertible matrix if $\left(A+I\right)^3=0$ and find $A^{-1}$ If $A\in M_{n\times n}\left(\mathbb{R}\right)$ is such that $ (A+I)^3=0$, show that $A$ is an invertible matrix and find the inverse of $A$.

My idea was:
\begin{eqnarray*}
0&amp;=&amp;\left(A+I\right)^3\\
&amp;=&amp;A^3+3A^2I+3AI^2+I^3\\
&amp;=&amp;A^3+3A^2+3A+I,
\end{eqnarray*}
then
$$I=-A^3-3A^2-3A,$$
so
$$I=A\left(-A^2-3A-3I\right).$$
It follows that
$$A^{-1}=-A^2-3A-3I.$$
Now I found the inverse matrix, but how does this show that an inverse actually exists?
Your idea is good, it works. You have shown that
$$A(-A^2-3A-3I)=I,$$
which means that $-A^2-3A-3I=A^{-1}$ by definition of the inverse. This also shows that the inverse exists, because you have constructed it already.
All of you are right actually but for one minor overlook perhaps... the given relationship between A and I cannot be interpreted as characterstic polynomial but rather MINIMAL POLYNOMIAL because A is of order n rather than three. Keeping that in mind, proving invertibility of A is simple: all eigenvalues of A are non-zero; each being -1, and therefore, det(A)= Product of eigenvalues is non-zero as well and hence, inverse of A exists. As far as, construction of the inverse goes, may be we require more information, just like that in this case A has been assumed to be a 3 x 3 matrix.