# Operations on Algebraic Fractions

• August 11th 2008, 11:50 AM
comet2000
Operations on Algebraic Fractions
Reduce each algebraic fraction.

$\frac{4-2u}
{u^3-8}$
• August 11th 2008, 11:59 AM
Quick
Quote:

Originally Posted by comet2000
Reduce each algebraic fraction.

$\frac{(4-2u)}
\frac{(u^3-8)}$

so is it $\frac{4-2u}{u^3-8}$?

I'll assume it is.

So let's rewrite things: $\frac{2(2-u)}{(u-2)(u^2+2u+2)}$

Let's move that negative: $\frac{2(2-u)}{-1(2-u)(u^2+2u+2)}$

Now we can cancel out the 2-u to get: $\frac{2}{-1(u^2+2u+2)}$

So therefore: $\frac{4-2u}{u^3-8}=\frac{-2}{u^2+2u+2}$