November 16th, 2016, 05:24 AM  #1 
Newbie Joined: Nov 2016 From: London Posts: 4 Thanks: 0  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 
November 16th, 2016, 05:44 AM  #2 
Math Team Joined: Jul 2011 From: Texas Posts: 2,094 Thanks: 972  $\dfrac{(2^{1/6} \cdot 6^{1/6}) \cdot (2^{1/3} \cdot 3^{1/3})}{(2^{1/3} \cdot 6^{1/6}) \cdot 2^{1/3}}$ can you finish the simplification? 
November 20th, 2016, 10:29 AM  #3 
Newbie Joined: Nov 2016 From: London Posts: 4 Thanks: 0 
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 11:14 AM. 
November 20th, 2016, 11:15 AM  #4 
Global Moderator Joined: Dec 2006 Posts: 16,180 Thanks: 1144 
It can be simplified further.

November 20th, 2016, 01:38 PM  #5 
Math Team Joined: May 2013 From: The Astral plane Posts: 1,437 Thanks: 528 Math Focus: Wibbly wobbly timeywimey stuff.  
November 21st, 2016, 03:06 AM  #6 
Newbie Joined: Nov 2016 From: London Posts: 4 Thanks: 0 
ooooohhhhhh!!! Of course! $\dfrac{(2^{1/6}) \cdot (3^{1/3})}{2^{1/3}}$ $=(2^{1/6}) \cdot (3^{1/3})$ or $\dfrac{3^{1/3}}{2^{1/6}}$ 
November 22nd, 2016, 01:08 AM  #7 
Newbie Joined: Nov 2016 From: London Posts: 4 Thanks: 0  Would the final answer be $\dfrac{3^{1/3}}{2^{1/6}}$?
Last edited by skipjack; November 27th, 2016 at 04:39 AM. 
November 27th, 2016, 03:13 AM  #8 
Newbie Joined: Nov 2016 From: St. Louis, Missouri Posts: 23 Thanks: 4 Math Focus: arithmetic, fractions 
I tried solving the equation until the 16 times 1/12 part threw me off.

November 27th, 2016, 04:47 AM  #9 
Global Moderator Joined: Dec 2006 Posts: 16,180 Thanks: 1144 
It's 16 raised to the power of 1/12, not 16 times 1/12.


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