Reduce each algebraic fraction.
$\displaystyle \frac{4-2u}
{u^3-8}$
so is it $\displaystyle \frac{4-2u}{u^3-8}$?
I'll assume it is.
So let's rewrite things: $\displaystyle \frac{2(2-u)}{(u-2)(u^2+2u+2)}$
Let's move that negative: $\displaystyle \frac{2(2-u)}{-1(2-u)(u^2+2u+2)}$
Now we can cancel out the 2-u to get: $\displaystyle \frac{2}{-1(u^2+2u+2)}$
So therefore: $\displaystyle \frac{4-2u}{u^3-8}=\frac{-2}{u^2+2u+2}$