Let $$A$$ and $$B$$ be two $$3 \times 3$$ real matrices such that $$(A^2 - B^2)$$ is invertible matrix. If $$A^5 = B^5$$ and $$A^3B^2 = A^2B^3$$, then the value of the determinant of the matrix $$A^3 + B^3$$ is equal to:

Given,

$$A^5=B^5$$

and

$$A^3B^2=A^2B^3$$

Also,

$$A^2-B^2$$

is invertible.

From

$$A^3B^2=A^2B^3$$

we get

$$A^2B^2(A-B)=0$$

Now,

$$A^5-B^5=0$$

Factorizing,

$$(A-B)(A^4+A^3B+A^2B^2+AB^3+B^4)=0$$

Since powers of matrices commute here, write

$$A^4+A^3B+A^2B^2+AB^3+B^4$$

as

$$(A^2+B^2)(A^2+AB+B^2)-A^2B^2$$

Using

$$A^2B^2(A-B)=0,$$

we get

$$A^2B^2=0$$

Hence,

$$A^5-B^5=(A-B)(A^2+B^2)(A^2+AB+B^2)=0$$

But

$$A^2-B^2=(A-B)(A+B)$$

is invertible.

Therefore,

$$A-B$$

is invertible.

Hence from

$$(A-B)(A^4+A^3B+A^2B^2+AB^3+B^4)=0$$

we get

$$A^4+A^3B+A^2B^2+AB^3+B^4=0$$

Now multiply by

$$A+B$$

Using algebraic identity,

$$A^5+B^5=0$$

But given

$$A^5=B^5$$

Therefore,

$$2A^5=0$$

$$A^5=0$$

Hence,

$$B^5=0$$

Thus both $$A$$ and $$B$$ are nilpotent matrices.

Therefore,

$$A^3+B^3$$

is also nilpotent.

Determinant of a nilpotent matrix is always zero.

Hence,

$$\boxed{0}$$