Given: x y z=9 and xy yz zx=23

To find: The value of [x^(3) y^(3) z^(3)-3xyz]

Solution:
We can use the identity (a+b+c)^3=a^3+b^3+c^3+3(a+b)(b+c)(c+a) to solve this question.

Let's calculate the value of (x+y+z)^3 first:
(x+y+z)^3 = x^3+y^3+z^3+3(x+y)(y+z)(z+x)
We know that x y z=9 and xy yz zx=23, so we can substitute these values in the above equation to get:
(x+y+z)^3 = x^3+y^3+z^3+3(9x+9y+9z-xyz)
(x+y+z)^3 = x^3+y^3+z^3+243-27xyz

Now, let's substitute the value of (x+y+z)^3 in the expression [x^(3) y^(3) z^(3)-3xyz] to get:
[x^(3) y^(3) z^(3)-3xyz] = x^3+y^3+z^3+243-27xyz-3xyz
[x^(3) y^(3) z^(3)-3xyz] = x^3+y^3+z^3-30xyz+243

We know that x y z=9, so xyz=729. Substituting this value in the above equation, we get:
[x^(3) y^(3) z^(3)-3xyz] = x^3+y^3+z^3-2187+243
[x^(3) y^(3) z^(3)-3xyz] = x^3+y^3+z^3-1944

Therefore, the value of [x^(3) y^(3) z^(3)-3xyz] is -1944.

Answer: [x^(3) y^(3) z^(3)-3xyz]=-1944.