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November 16th, 2016, 04:24 AM   #1
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help with a question

Hi all,

I've been set a question and was wondering if I was doing it correctly, here it is:

Maths unit 1 assignment 3 question 1 b.jpg

My instinct is telling me to factorise the numbers individually but from there I'm a little lost how to simplify it further.

Any help or assistance would be massively appreciated!!

Cheers all,
D
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November 16th, 2016, 04:44 AM   #2
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$\dfrac{(2^{1/6} \cdot 6^{1/6}) \cdot (2^{1/3} \cdot 3^{1/3})}{(2^{1/3} \cdot 6^{1/6}) \cdot 2^{1/3}}$

can you finish the simplification?
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November 20th, 2016, 09:29 AM   #3
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Hi Skeeter, thanks for the reply,

Been thinking about this for a wee bit, and I've come to this:

$\dfrac{(2^{1/6}) \cdot (3^{1/3})}{2^{1/3}}$

Last edited by skipjack; November 20th, 2016 at 10:14 AM.
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November 20th, 2016, 10:15 AM   #4
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It can be simplified further.
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November 20th, 2016, 12:38 PM   #5
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Quote:
Originally Posted by Alavanca11 View Post
Hi Skeeter, thanks for the reply,

Been thinking about this for a wee bit, and I've come to this:

$\dfrac{(2^{1/6}) \cdot (3^{1/3})}{2^{1/3}}$
If I tell you that $\displaystyle \frac{x^a}{x^b} = x^{a - b}$, what can you do with it?

-Dan
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November 21st, 2016, 02:06 AM   #6
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ooooohhhhhh!!! Of course!

$\dfrac{(2^{1/6}) \cdot (3^{1/3})}{2^{1/3}}$ $=(2^{-1/6}) \cdot (3^{1/3})$

or $\dfrac{3^{1/3}}{2^{1/6}}$
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November 22nd, 2016, 12:08 AM   #7
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Quote:
Originally Posted by topsquark View Post
If I tell you that $\displaystyle \frac{x^a}{x^b} = x^{a - b}$, what can you do with it?

-Dan
Would the final answer be $\dfrac{3^{1/3}}{2^{1/6}}$?

Last edited by skipjack; November 27th, 2016 at 03:39 AM.
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November 27th, 2016, 02:13 AM   #8
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I tried solving the equation until the 16 times 1/12 part threw me off.
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November 27th, 2016, 03:47 AM   #9
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It's 16 raised to the power of 1/12, not 16 times 1/12.
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