Given x+y+z=6, xy+yz+zx=11, and x3+y3+z3=36, find the values of x, y, z.

The given system of equations is:

The provided identity is:

Substitute the known values into the identity:

Simplify the equation from Step 1:

Subtract 36 from both sides:

Divide by -3:

For a cubic polynomial with roots $x, y, z$, the general form is:

Using the values we have:
$x+y+z = 6$
$xy+yz+zx = 11$
$xyz = 6$
Substitute these values into the polynomial form:

We need to find the roots of $t^3 - 6t^2 + 11t - 6 = 0$.
By testing integer divisors of the constant term -6 (i.e., $\pm 1, \pm 2, \pm 3, \pm 6$):
For $t=1$: $1^3 - 6(1)^2 + 11(1) - 6 = 1 - 6 + 11 - 6 = 0$. So $t=1$ is a root.
Since $t=1$ is a root, $(t-1)$ is a factor of the polynomial. We can perform polynomial division or synthetic division.
Using synthetic division with root 1:

1 | 1  -6   11  -6
  |    1   -5    6
  -----------------
    1  -5    6    0

The quotient is $t^2 - 5t + 6$.
Now, we solve the quadratic equation $t^2 - 5t + 6 = 0$.
This factors as $(t-2)(t-3) = 0$.
The roots are $t=2$ and $t=3$.
Thus, the roots of the polynomial are $1, 2, 3$.

The real numbers $x, y, z$ are the roots of this polynomial. Since the system is symmetric, the values of $x, y, z$ are $1, 2, 3$ in any order.

The values for $x, y, z$ are $\boxed{1, 2, 3}$.
The cubic polynomial is $\boxed{t^3 - 6t^2 + 11t - 6 = 0}$.