We are given the matrix equation $A^3 - 3A + 6I = 0$, where $A$ is a square matrix and $I$ is the identity matrix.

To find $A^{-1}$, follow these steps:

1. Rewrite the equation as:

   $A^3 - 3A = -6I$

   Divide both sides by $-6$ to get:

   $\frac{A^3 - 3A}{-6} = I$

2. Now factor the left-hand side:

   $A(A^2 - 3I) = -6I$

   We want to express $A^{-1}$ in terms of $A$. Rewrite the equation to solve for $A^{-1}$:

   $A(A^2 - 3I) = -6I \quad \Rightarrow \quad A^{-1} = -\frac{1}{6}(A^2 - 3I)$

Thus, the inverse of the matrix $A$ is:

$A^{-1} = -\frac{1}{6}(A^2 - 3I)$

Would you like any further details or have any questions?

Here are some related questions for further exploration:

1. How can we check if a matrix is invertible?
2. What properties must a matrix have for an inverse to exist?
3. How would you find the inverse of a 2x2 matrix by hand?
4. What is the Cayley-Hamilton theorem and how does it apply to matrices?
5. How can this matrix equation help in finding higher powers of $A$?

Tip: When dealing with matrix equations involving inverses, always ensure the matrix is non-singular (determinant is non-zero) to guarantee the existence of $A^{-1}$.