Determine the number of possible values for the determinant of $A$, given that $A$ is a $n\times n$ matrix with real entries such that $A^3 - A^2 - 3A + 2I = 0$, where $I$ is the identity and $0$ is the all-zero matrix.
1. **Given the matrix equation**:
   \[
   A^3 - A^2 - 3A + 2I = 0
   \]
   We can factorize the polynomial on the left-hand side:
   \[
   A^3 - A^2 - 3A + 2I = (A - I)(A - 2I)(A + I) = 0
   \]
   This implies that the eigenvalues of \(A\) must satisfy one of the equations:
   \[
   \lambda - 1 = 0 \quad \text{or} \quad \lambda - 2 = 0 \quad \text{or} \quad \lambda + 1 = 0
   \]
   Therefore, the eigenvalues of \(A\) are \(\lambda = 1, 2, -1\).

2. **Diagonalizability of \(A\)**:
   Since \(A\) satisfies a polynomial equation with distinct roots, \(A\) is diagonalizable. This means \(A\) can be written as:
   \[
   A = PDP^{-1}
   \]
   where \(D\) is a diagonal matrix with the eigenvalues \(1, 2, -1\) on the diagonal.

3. **Determinant of \(A\)**:
   The determinant of \(A\) is the product of its eigenvalues. Let \(\alpha\), \(\beta\), and \(\gamma\) be the multiplicities of the eigenvalues \(2\), \(\frac{-1 - \sqrt{5}}{2}\), and \(\frac{-1 + \sqrt{5}}{2}\) respectively. Then:
   \[
   \alpha + \beta + \gamma = n
   \]
   The determinant of \(A\) is given by:
   \[
   \det(A) = 2^{\alpha} \left(\frac{-1 - \sqrt{5}}{2}\right)^{\beta} \left(\frac{-1 + \sqrt{5}}{2}\right)^{\gamma}
   \]

4. **Uniqueness of the function**:
   The function \((\beta, \gamma) \in \mathbb{N}^2 \rightarrow 2^{n - \beta - \gamma} \left(\frac{-1 - \sqrt{5}}{2}\right)^{\beta} \left(\frac{-1 + \sqrt{5}}{2}\right)^{\gamma}\) is one-to-one. To prove this, assume:
   \[
   \left(\frac{-1 - \sqrt{5}}{2}\right)^{\beta} \left(\frac{-1 + \sqrt{5}}{2}\right)^{\gamma} = 2^{\beta + \gamma}
   \]
   Since \(\left(\frac{-1 - \sqrt{5}}{2}\right) \left(\frac{-1 + \sqrt{5}}{2}\right) = -1\), we have:
   \[
   \left| \left(\frac{-1 - \sqrt{5}}{2}\right)^{\beta} \left(\frac{-1 + \sqrt{5}}{2}\right)^{\gamma} \right| = \left| (-1)^{\beta} \left(\frac{-1 + \sqrt{5}}{2}\right)^{\gamma - \beta} \right|
   \]
   This number is rational only if \(\beta = \gamma\). Therefore:
   \[
   (-1)^{\beta} = 2^{2\beta}
   \]
   This implies \(\beta = \gamma = 0\). Hence, the function is one-to-one.

5. **Counting the possible values**:
   The number of possible values for \(\det(A)\) is given by the number of non-negative integer solutions to \(\alpha + \beta + \gamma = n\). This is a classic combinatorial problem, and the number of solutions is given by:
   \[
   \binom{n + 2}{2} = \frac{(n + 2)(n + 1)}{2}
   \]

The final answer is \(\boxed{\frac{(n + 2)(n + 1)}{2}}\).