Saudi Arabian Mathematical Competitions

The system is equivalent to
$$\left\{ \begin{array}{l} x + y + z = -(w + 1) \\ x y + y z + z x = -\left(w^{2} - w + 1\right) \\ x y z = w^{3} - 1 \end{array} \right.$$
Multiplying the first two equations of the system we get
$$(x + y + z)(x y + y z + z x) = w^{3} + 1.$$
Eliminating $w$ between this relation and the last equation, it follows
$$(x + y + z)(x y + y z + z x) - x y z = 2.$$
The equation (1) is equivalent to
$$(x + y)(y + z)(z + x) = 2.$$
Taking into account the symmetry of equation (3) we have to consider only the following situations:
$$\left\{ \begin{array}{l} x + y = 2 \\ y + z = -1 \\ z + x = -1 \end{array} \quad \left\{ \begin{array}{l} x + y = 2 \\ y + z = 1 \\ z + x = 1 \end{array} \quad \left\{ \begin{array}{l} x + y = -2 \\ y + z = 1 \\ z + x = -1 \end{array} \right. \right. \right.$$
Because $w^{2} - w + 1 > 0$, from the second equation of (1) it follows that the solutions to the above system must satisfy
$$x y + y z + z x < 0.$$
The first system has solution $(x, y, z) = (1, 1, -2)$. Replacing in the first equation of (1) we obtain $w = -1$. In this case we get solutions $(1, 1, -2, -1)$, $(1, -2, 1, -1)$, $(-2, 1, 1, -1)$.
The solutions to the second and third system do not satisfy condition $x y + y z + z x < 0$, hence they will not give solutions to our problem.

Solution 2:
$x, y, z$ are the roots of the polynomial
$$P(t) = t^{3} + (w + 1) t^{2} - \left(w^{2} - w + 1\right) t - w^{3} + 1$$
We have
$$P(t) = (t + w + 1)\left(t^{2} - w^{2} + w - 1\right) + 2.$$
Since $P(x) = 0$, it follows
$$(x + w + 1)\left(x^{2} - w^{2} + w - 1\right) = -2$$
There are four possibilities
$$\left\{ \begin{array}{l} x + w + 1 = 1 \\ x^{2} - w^{2} + w - 1 = -2 \end{array} \quad \left\{ \begin{array}{l} x + w + 1 = -1 \\ x^{2} - w^{2} + w - 1 = 2 \end{array} \right. \right.$$
$$\left\{ \begin{array}{l} x + w + 1 = 2 \\ x^{2} - w^{2} + w - 1 = -1 \end{array} \quad \left\{ \begin{array}{l} x + w + 1 = -2 \\ x^{2} - w^{2} + w - 1 = 1 \end{array} \right. \right.$$
The second and the fourth system have no solutions in integers. From the first and from the second system we get $w = -1$ and $w = 1$.
If $w = -1$, the polynomial $P$ is $P(t) = t^{3} - 3 t + 2$ with roots $x = y = 1, z = -2$. We get solutions $(1, 1, -2, -1)$, $(1, -2, 1, -1)$, $(-2, 1, 1, -1)$.
If $w = 1$, the polynomial $P$ is $P(t) = t^{3} + 2 t^{2} - t$ and not all the roots are integers.