Roots of Unity and Cubic Equations

I was doing a question and i did the first part fine where i had to show that:

(X+Y+Z)(X+wY+[w^2]Z)(X+[w^2]Y+wZ)= X^3+Y^3+Z^3-3XYZ.

where w=exp(i*2pi/3) i.e. the primitive cube root of unity.

Now the second part implies that the above should be used to solve the equatoion:

x^3-6x+6=0. (*)

My first thoughts were to get the simplified expression in the first part to match equation (*) with X essentially representing x. And then you would have roots as we have shown the expression can be rewritten in a fully factorised form..i.e. the roots are easily deduced from this.

However that means we require numbers for Y,Z such that:

YZ=2 and Y^3+Z^3=6

And i cant think that any values will satisfy this? Can anyone help

Substituting Z = 2/Y into Y^3 + Z^3 = 6 gives a quadratic in Y^3. One possible pair is Y = 2^(1/3), Z = 4^(1/3).