If $(10 - x)(x - 2) = 11$, then compute $(10 - x)^3 + (x - 2)^3$.
We expand $(10 - x)^3 + (x - 2)^3$ as a sum of cubes, which states that \[(a + b)(a^2 - ab + b^2) = a^3 + b^3.\]Thus, let $a = 10 - x$ and $b = x - 2$. Then $(10 - x)^3 + (x - 2)^3 = (10 - x + x - 2)((10 - x)^2 - (10 - x)(x - 2) + (x - 2)^2)$.
Now, since $(10 - x)(x - 2) = 11$, we know that \[(10 - x + x - 2)((10 - x)^2 - (10 - x)(x - 2) + (x - 2)^2) = 8((10 - x)^2 - 11 + (x - 2)^2).\]But, notice that $(10 - x)^2 - 11 + (x - 2)^2$ is equal to \[(x^2 - 20x + 100) - 11 + (x^2 - 4x + 4) = 2x^2 - 24x + 93.\]Therefore, $8((10 - x)^2 - 11 + (x - 2)^2) = 8 \cdot (2x^2 - 24x + 93) = \boxed{16x^2 - 192x + 744}$.
To solve the equation $(10 - x)(x - 2) = 11$, let's expand the left side:
$(10 - x)(x - 2) = 11$
$10x - 20 - x^2 + 2x = 11$
Simplifying this equation, we can rewrite it as:
$10x - x^2 - 18 = 0$
Now, let's solve this quadratic equation. We can rearrange the terms to get:
$x^2 - 10x + 18 = 0$
Using the quadratic formula, $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, where $a=1$, $b=-10$, and $c=18$, we can solve for $x$:
$x = \frac{-(-10) \pm \sqrt{(-10)^2 - 4(1)(18)}}{2(1)}$
$x = \frac{10 \pm \sqrt{100 - 72}}{2}$
$x = \frac{10 \pm \sqrt{28}}{2}$
$x = \frac{10 \pm 2\sqrt{7}}{2}$
$x = 5 \pm \sqrt{7}$
Now that we have found the values of $x$, we can compute $(10 - x)^3 + (x - 2)^3$ for each value:
For $x = 5 + \sqrt{7}$:
$(10 - (5 + \sqrt{7}))^3 + ((5 + \sqrt{7}) - 2)^3$
$(5 - \sqrt{7})^3 + (3 + \sqrt{7})^3$
For $x = 5 - \sqrt{7}$:
$(10 - (5 - \sqrt{7}))^3 + ((5 - \sqrt{7}) - 2)^3$
$(5 + \sqrt{7})^3 + (3 - \sqrt{7})^3$
These expressions require further simplification and expansion to compute their values.