# Dividing with Brackets And Indicies

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• May 11th 2008, 05:26 AM
mcaelli
Dividing with Brackets And Indicies
My Question is this:

(12p²q-³r)-³ / 144q^7
-------------_________
----------------5p²

Note: Last part is 144q^7 OVER 5p², not all over 5p²

I have attempeted the first part and ended up with -1728p^4q^-9r³.
Am i on the right track here? If so where do i go from here? How do i do the 144q^7 over 5p², and then incorporate it back into the -1728... part?
• May 11th 2008, 05:36 AM
topsquark
Quote:

Originally Posted by mcaelli
My Question is this:

(12p²q-³r)-³ / 144q^7
-------------_________
----------------5p²

Note: Last part is 144q^7 OVER 5p², not all over 5p²

I have attempeted the first part and ended up with -1728p^4q^-9r³.
Am i on the right track here? If so where do i go from here? How do i do the 144q^7 over 5p², and then incorporate it back into the -1728... part?

If I am reading this right:
$\frac{(12p^2q - 3r)^{-3}}{\frac{144q^7}{5p^2}}$

Get rid of the fraction in the denominator:
$= \frac{(12p^2q - 3r)^{-3}}{\frac{144q^7}{5p^2}} \cdot \frac{5p^2}{5p^2}$

$= \frac{5p^2(12p^2q - 3r)^{-3}}{144q^7}$

Now take care of the negative exponent:
$= \frac{5p^2}{144q^7(12p^2q - 3r)^3}$

-Dan
• May 11th 2008, 05:43 AM
mcaelli
Ah the first bit with the brackets is : (144p^2 q^-3 r)^-3.
Does this alter what you have done very much?
• May 11th 2008, 06:59 AM
Soroban
Hello, mcaelli!

Quote:

Simplify: . $\frac{(12p^2q^{-3}r)^{-3}}{\frac{144q^7}{5p^2}}$

We have: . $\frac{12^{-3}(p^2)^{-3}(q^{-3})^{-3}r^{-3}}{1} \cdot \frac{5p^2}{144q^7} \;=\;\frac{12^{-3}p^{-6}q^9r^{-3}}{1} \cdot\frac{5p^2}{144q^7}$

. . $= \;\frac{q^9}{12^3p^6r^3}\cdot\frac{5p^2}{12^2q^7} \;=\;\frac{5p^2q^9}{12^5p^6q^7r^3} \;=\;\frac{5q^2}{12^5p^4r^3}$