Operations on Algebraic Fractions

**comet2000** · August 11th 2008, 11:50 AM

Reduce each algebraic fraction.

$\frac{4-2u}<br /> {u^3-8}$

**Quick** · August 11th 2008, 11:59 AM

Originally Posted by comet2000

Reduce each algebraic fraction.

$\frac{(4-2u)}<br /> \frac{(u^3-8)}$

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

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