Given:
- M is a 3x3 real matrix
- M^3 = M (M raised to the power of 3 is equal to M)

To prove:
I. M is invertible
II. Eigen values of M are distinct
III. M is singular

Proof:

I. M is invertible:
- We need to prove that the determinant of M is not equal to zero.
- Let's assume that M is not invertible, which means the determinant of M is equal to zero.
- Then det(M) = 0
- We can rewrite the equation M^3 = M as M^3 - M = 0
- Factoring out M from the equation, we get M(M^2 - I) = 0, where I is the identity matrix.
- Since M^2 - I is a 3x3 matrix, if M is not invertible, then M^2 - I must be the zero matrix.
- But M^2 - I = 0 implies M^2 = I, which contradicts the given condition M^3 = M.
- Hence, our assumption that M is not invertible is incorrect.
- Therefore, M is invertible.

II. Eigen values of M are distinct:
- We need to prove that all eigenvalues of M are distinct.
- Let's assume that M has two eigenvalues, λ1 and λ2, where λ1 ≠ λ2.
- By the definition of eigenvalues, we have Mv1 = λ1v1 and Mv2 = λ2v2, where v1 and v2 are eigenvectors corresponding to the eigenvalues λ1 and λ2 respectively.
- Taking the cube of both equations, we get M^3v1 = λ1^3v1 and M^3v2 = λ2^3v2.
- Since M^3 = M, we can rewrite the equations as Mv1 = λ1^3v1 and Mv2 = λ2^3v2.
- But λ1^3v1 = λ1v1 and λ2^3v2 = λ2v2, which implies λ1^3 = λ1 and λ2^3 = λ2.
- The only real numbers that satisfy this condition are 0, 1, and -1.
- Therefore, the eigenvalues of M can only be 0, 1, or -1.
- Since λ1 ≠ λ2, we conclude that the eigenvalues of M are distinct.

III. M is singular:
- We need to prove that M is singular, which means the determinant of M is equal to zero.
- From part I, we have already proven that M is invertible, which implies the determinant of M is not equal to zero.
- Therefore, our assumption that M is singular is incorrect.

Conclusion:
- From the above proof, we can conclude that:
I. M is invertible
II. Eigenvalues of M are distinct
III. M is not singular