Let A = [ cases x & y & z y & z & x z & x & y cases ], where x, y...

$⇒ A{A}^{'} = I$ (as ${A}^{'} = A$)

$⇒ A$ is orthogonal

So, ${x}^{2} + {y}^{2} + {z}^{2} = 1$ and $xy + yz + zx = 0$

$⇒ {\left(x + y + z\right)}^{2} = 1 + 2 \times 0$

$⇒ x + y + z = 1$

Thus,

${x}^{3} + {y}^{3} + {z}^{3} = 3 \times 2 + 1 \times \left(1 - 0\right)$

$= 7$