Let $$A = \begin{bmatrix} x & y & z \\ y & z & x \\ z & x & y \end{bmatrix}$$, where $$x, y$$ and $$z$$ are real numbers such that $$x + y + z > 0$$ and $$xyz = 2$$. If $$A^2 = I_3$$, then the value of $$x^3 + y^3 + z^3$$ is ______

The matrix $$A = \begin{bmatrix} x & y & z \\ y & z & x \\ z & x & y \end{bmatrix}$$ is a circulant matrix. Since $$A^2 = I_3$$, we have $$A^2 = I$$, meaning $$A$$ is an involutory matrix, so its eigenvalues are $$\pm 1$$.

For a circulant matrix with first row $$(x, y, z)$$, the eigenvalues are $$\lambda_k = x + y\omega^k + z\omega^{2k}$$ where $$\omega = e^{2\pi i/3}$$ for $$k = 0, 1, 2$$.

The eigenvalue for $$k = 0$$ is $$\lambda_0 = x + y + z$$. Since $$x + y + z > 0$$ and $$\lambda_0 = \pm 1$$, we must have $$x + y + z = 1$$.

Since $$A^2 = I_3$$, we also know $$\det(A)^2 = 1$$, so $$\det(A) = \pm 1$$. The determinant of this circulant matrix is $$x^3 + y^3 + z^3 - 3xyz$$. Since $$xyz = 2$$, we get $$\det(A) = x^3 + y^3 + z^3 - 6$$.

Now using the identity $$x^3 + y^3 + z^3 - 3xyz = (x + y + z)(x^2 + y^2 + z^2 - xy - yz - zx)$$, we have $$\det(A) = 1 \cdot (x^2 + y^2 + z^2 - xy - yz - zx)$$.

Also, $$A^2 = I$$ means $$AA = I$$. Computing the $$(1,1)$$ entry of $$A^2$$: $$x^2 + y^2 + z^2 = 1$$. Computing the $$(1,2)$$ entry: $$xy + yz + zx = 0$$.

Therefore, $$\det(A) = 1 - 0 = 1$$, and $$x^3 + y^3 + z^3 - 6 = 1$$, giving $$x^3 + y^3 + z^3 = 7$$.