$M$ and $N$ are real unequal $n\times n$ matrices satisfying $M^3=N^3$ and $M^2N=N^2M$. Can we choose $M$ and $N$ so that $M^2+N^2$ is invertible?
1. Given that \( M \) and \( N \) are real unequal \( n \times n \) matrices satisfying \( M^3 = N^3 \) and \( M^2N = N^2M \), we need to determine if \( M^2 + N^2 \) can be invertible.

2. Let's start by analyzing the given conditions:
   - \( M^3 = N^3 \)
   - \( M^2N = N^2M \)

3. From \( M^3 = N^3 \), we can write:
   \[
   M^3 - N^3 = 0
   \]
   Using the factorization of the difference of cubes, we get:
   \[
   (M - N)(M^2 + MN + N^2) = 0
   \]

4. Since \( M \) and \( N \) are unequal, \( M \neq N \). Therefore, \( M - N \) is not the zero matrix. This implies:
   \[
   M^2 + MN + N^2 = 0
   \]

5. Next, consider the second condition \( M^2N = N^2M \). We can rewrite this as:
   \[
   M^2N - N^2M = 0
   \]
   This can be factored as:
   \[
   M^2N - N^2M = M^2N - MNM + MNM - N^2M = M(MN - NM) + (MN - NM)M = 0
   \]
   Since \( M \) and \( N \) are not necessarily commutative, this does not provide additional information directly.

6. Now, we need to determine if \( M^2 + N^2 \) can be invertible. Suppose \( M^2 + N^2 \) is invertible. Then, its determinant must be non-zero:
   \[
   \det(M^2 + N^2) \neq 0
   \]

7. However, from step 4, we have:
   \[
   M^2 + MN + N^2 = 0 \implies M^2 + N^2 = -MN
   \]
   If \( M^2 + N^2 \) is invertible, then \( -MN \) must also be invertible. This implies that both \( M \) and \( N \) must be invertible matrices.

8. If \( M \) and \( N \) are invertible, then we can consider the product \( MN \). Since \( M^2 + N^2 = -MN \), we have:
   \[
   \det(M^2 + N^2) = \det(-MN) = (-1)^n \det(M) \det(N)
   \]
   For \( M^2 + N^2 \) to be invertible, \( \det(M) \det(N) \neq 0 \), which means both \( \det(M) \) and \( \det(N) \) are non-zero.

9. Therefore, it is possible to choose \( M \) and \( N \) such that \( M^2 + N^2 \) is invertible, provided that \( M \) and \( N \) are invertible matrices themselves.

The final answer is \( \boxed{ M^2 + N^2 } \) can be invertible if \( M \) and \( N \) are chosen to be invertible matrices.