If x = 2, y = 1, z = -3 then x 3 + y 3 + z 3 - 3xyz equals

Let's look at the algebraic identity for the given expression:
$$x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2-xy-yz-zx)$$
First, let's calculate the sum of the variables $(x+y+z)$ using the given values:
$$x+y+z=2+1+(-3)$$
$$x+y+z=3-3=0$$
Since the term $(x+y+z)=0$, substituting this into our identity simplifies the entire product to zero:
$$x^3+y^3+z^3-3xyz=(0)\times (x^2+y^2+z^2-xy-yz-zx)=0$$