Compute $\frac{x^6-16x^3+64}{x^3-8}$ when $x=6$.
I wonder if there is a way to simplify the fraction before plugging in $x=6$. I notice that the numerator is a quadratic function of $x^3$, and the denominator is a linear function of $x^3$. This suggests that I can try to factor the numerator as a product of two binomials of the form $(x^3-a)(x^3-b)$, and see if the denominator is one of the factors. To do this, I need to find two numbers $a$ and $b$ that add up to $-16$ and multiply to $64$. I can use trial and error to find such numbers. One pair that works is $a=-8$ and $b=-8$. So the numerator can be factored as $(x^3-8)(x^3-8)$. Now I see that the denominator is exactly one of the factors, so I can cancel them out. This leaves me with $\frac{x^3-8}{1}$, which is just $x^3-8$. Now I can plug in $x=6$ and evaluate the expression. I get $6^3-8=216-8=208$. 