Simplifying fractions with exponents

• Nov 2nd 2006, 02:36 PM
violinist
Simplifying fractions with exponents
It should be simple, but I keep messing it up. I really need a step-by-step description of the process if it is possible, as the problem lies in my messing up one of the steps, I imagine, but I won't know which one(s) unless I see the correct steps.

Now, for the problem.
(-2xyz)-³
(x²z-³)-³

The instructions are "Simplify".

Thanks sooooo much for any help!

P.S., I am homeschooled, and have the answer key. After checking the answer I realized that I have the numbers for the exponents correct, but I am somehow doing the coefficients wrong. Somewhere they are getting an "8x to the blah blah", and I have no idea how they got an 8?
• Nov 2nd 2006, 02:51 PM
CaptainBlack
Quote:

Originally Posted by violinist
It should be simple, but I keep messing it up. I really need a step-by-step description of the process if it is possible, as the problem lies in my messing up one of the steps, I imagine, but I won't know which one(s) unless I see the correct steps.

Now, for the problem.
(-2xyz)-³
(x²z-³)-³

The instructions are "Simplify".

Thanks sooooo much for any help!

P.S., I am homeschooled, and have the answer key. After checking the answer I realized that I have the numbers for the exponents correct, but I am somehow doing the coefficients wrong. Somewhere they are getting an "8x to the blah blah", and I have no idea how they got an 8?

I will focus on the numerical coefficient

We have:

$
\frac{(-2x y z)^{-3}}{(x^2z^{-3})^{-3}}
=(-2)^{-3}\frac{(x y z)^{-3}}{(x^2z^{-3})^{-3}}
=\frac{1}{(-2)^3}\frac{(x y z)^{-3}}{(x^2z^{-3})^{-3}}
$

Now $(-2)^3=(-1)^3\ 2^3=-(2^3)=-8$ hence:

$
\frac{(-2x y z)^{-3}}{(x^2z^{-3})^{-3}}
=-\frac{1}{8}\ \frac{(x y z)^{-3}}{(x^2z^{-3})^{-3}}
$

RonL