If $x+y+z=11$, $xy+yz+zx=-6$, and $x^3+y^3+z^3=1,604$, then the value of $xyz$ is:

The correct answer is 25.

To find the value of $xyz$, we can use standard algebraic identities.

We are given the following values:
1. $x+y+z=11$
2. $xy+yz+zx=-6$
3. $x^3+y^3+z^3=1604$

First, recall the key algebraic identity relating the sum of cubes to the product of terms:
$x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2-(xy+yz+zx\left)\right)$

To evaluate the right-hand side of this identity, we first need to determine the value of $x^2+y^2+z^2$. We can find this by squaring the sum $x+y+z$:
${(x+y+z)}^2=x^2+y^2+z^2+2(xy+yz+zx)$

Rearranging this formula to solve for $x^2+y^2+z^2$ gives:
$x^2+y^2+z^2={(x+y+z)}^2-2(xy+yz+zx)$

Substitute the given values into the equation:
$x^2+y^2+z^2={\left(11\right)}^2-2(-6)$
$x^2+y^2+z^2=121+12$
$x^2+y^2+z^2=133$

Now, substitute $x^2+y^2+z^2=133$ and the other known values back into our main identity:
$1604-3xyz=\left(11\right)(133-(-6\left)\right)$
$1604-3xyz=11(133+6)$
$1604-3xyz=11\left(139\right)$
$1604-3xyz=1529$

Rearrange the terms to solve for $3xyz$:
$3xyz=1604-1529$
$3xyz=75$

Divide by 3 to find the final value of $xyz$:
$xyz=25$