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October 26th, 2014, 03:53 PM  #1 
Newbie Joined: Oct 2014 From: Naperville, IL Posts: 5 Thanks: 0  I am a bit confused about a simple problem
I don't understand how the answer to this question was obtained. Why does the Cubed Root of the Square root of 27 = The square root of three? Can anyone help? 
October 26th, 2014, 04:04 PM  #2 
Math Team Joined: Apr 2010 Posts: 2,780 Thanks: 361 
In symbols, it says: $\displaystyle \sqrt[3]{\sqrt{27}} = \sqrt{3}$. Now, $\displaystyle \sqrt[3]{\sqrt{27}} = \sqrt{\sqrt[3]{27}} $. Can you finish? 
October 26th, 2014, 04:05 PM  #3 
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada  The Forest City Posts: 7,950 Thanks: 1141 Math Focus: Elementary mathematics and beyond 
$\displaystyle (27^{1/2})^{1/3}=27^{1/6}=(27^{1/3})^{1/2}=3^{1/2}$

October 27th, 2014, 02:31 AM  #4  
Senior Member Joined: Jun 2013 From: London, England Posts: 1,316 Thanks: 116  Quote:
aaaaaa = (aaa)(aaa) = (square root of 27)(square rootof 27) = 27 Let b = square root of 3 bbbbbb = (bb)(bb)(bb) = (3)(3)(3) = 27 Therefore a = b  
October 27th, 2014, 02:43 AM  #5  
Math Team Joined: Apr 2010 Posts: 2,780 Thanks: 361  Quote:
Let a = cubed root of square root of 27 aaaaaa = (aaa)(aaa) = (square root of 27)(square rootof 27) = 27 Let b = square root of 3 bbbbbb = (bb)(bb)(bb) = (3)(3)(3) = 27 Therefore a = b  
October 27th, 2014, 02:58 AM  #6  
Senior Member Joined: Jun 2013 From: London, England Posts: 1,316 Thanks: 116  Quote:
aaaaaa = (aaa)(aaa) = (square root of 27)(square rootof 27) = 27 Let b = square root of 3 bbbbbb = (bb)(bb)(bb) = (3)(3)(3) = 27 Therefore a = b (as a & b are positive)  
October 27th, 2014, 08:06 AM  #7 
Math Team Joined: Apr 2010 Posts: 2,780 Thanks: 361 
Raising to an (even) power doesn't necessarily create extraneous nonpositive solutions, like for example squaring both sides of x2 = 1.

October 27th, 2014, 09:20 AM  #8  
Senior Member Joined: Jun 2013 From: London, England Posts: 1,316 Thanks: 116  Quote:
If a & b are defined as square or cube roots, then they are implicitly positive. So, there was nothing wrong in my original post, although it was better to point out in the conclusion that I was using the fact that a, b > 0.  
October 28th, 2014, 03:08 AM  #9 
Math Team Joined: Apr 2010 Posts: 2,780 Thanks: 361 
I didn't try to point out that you were wrong, just that one has to be more careful. Few quibbles: a,b >0 then a^6=b^6 ⇒ a=b. If a & b are defined as square roots, then they are implicitly positive. In the exercise, both sides are positive as all terms have a square root. I ask to be careful for the naive following work out: Solve x  2 = 1 Square: (x  2)^2 = 1^2 = 1 So x  2 = 1 or x  2 = 1 and x = 3 or x = 1. As x = 3 and x = 1 are both positive, both are solutions. Yes, this is extremely naive and wrong and I might be pushing this too far. 

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