Simplifying Fraction

**maca404** · Mar 16th 2011, 04:14 PM

$\frac{(2ab)^-2}{2(ab)^3}$

Can this fraction be simlified ?

I expanded it out to:

$\frac{2a^-2 + b^-2}{2a^3 + 2b^3}$

Even in this form I can't see common factors in the numerator so does that mean this cant be simplified ?

**e^(i*pi)** · Mar 16th 2011, 04:21 PM

Where did those + signs come from? The law is $(ab)^n = a^nb^n$

Can you rewrite your fraction using that rule?

**maca404** · Mar 16th 2011, 04:35 PM

Originally Posted by e^(i*pi)

Where did those + signs come from? The law is $(ab)^n = a^nb^n$

Can you rewrite your fraction using that rule?

Ok so

$\frac{2a^-2 b^-2}{2a^3 2b^3}$

Does this mean the a^-2 and a^3 be canceled ? Thats where im running in to trouble.

**skeeter** · Mar 16th 2011, 05:24 PM

$\dfrac{(2ab)^{-2}}{2(ab)^3} = \dfrac{1}{(2ab)^2 \cdot 2(ab)^3} =$ (?)

finish it

**maca404** · Mar 16th 2011, 05:55 PM

$\frac{1}{2*2*2a^2*b^2*a^3*b^3}$

$\frac{1}{8a^2*b^2*a^3*b^3}$

$\frac{1}{8a^2b^2*a^3*b^3}$

$\frac{1}{8a^5b^5}$

Am I there ?

**Wilmer** · Mar 16th 2011, 11:10 PM

"There" you are!

**maca404** · Mar 20th 2011, 06:17 PM

In the fraction above we can move (2ab)^-2 down with the denominator and change the power, I understand that is just what you have to do but I am wondering why does this work and when can we apply this method and when not ?

**skeeter** · Mar 20th 2011, 06:57 PM

Originally Posted by maca404

In the fraction above we can move (2ab)^-2 down with the denominator and change the power, I understand that is just what you have to do but I am wondering why does this work and when can we apply this method and when not ?

why it works ...

you should know that $\dfrac{x^a}{x^b} = x^{a-b}$

so , what do you get if $a = 0$ ?

now you have the rule ...

$x^{-b} = \dfrac{1}{x^b}$

so ...

$(2ab)^{-2} = \dfrac{1}{(2ab)^2}$

... all day long.

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