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October 26th, 2014, 04:53 PM   #1
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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?
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October 26th, 2014, 05:04 PM   #2
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In symbols, it says: $\displaystyle \sqrt[3]{\sqrt{27}} = \sqrt{3}$. Now,
$\displaystyle \sqrt[3]{\sqrt{27}} = \sqrt{\sqrt[3]{27}} $. Can you finish?
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October 26th, 2014, 05:05 PM   #3
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$\displaystyle (27^{1/2})^{1/3}=27^{1/6}=(27^{1/3})^{1/2}=3^{1/2}$
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October 27th, 2014, 03:31 AM   #4
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Quote:
Originally Posted by ttryon View Post
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?
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
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October 27th, 2014, 03:43 AM   #5
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Quote:
Originally Posted by Pero View Post
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
Careful there, similarily,
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
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October 27th, 2014, 03:58 AM   #6
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Quote:
Originally Posted by ttryon View Post
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?
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 (as a & b are positive)
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October 27th, 2014, 09:06 AM   #7
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Raising to an (even) power doesn't necessarily create extraneous non-positive solutions, like for example squaring both sides of x-2 = 1.
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October 27th, 2014, 10:20 AM   #8
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Quote:
Originally Posted by Hoempa View Post
Raising to an (even) power doesn't necessarily create extraneous non-positive solutions, like for example squaring both sides of x-2 = 1.
$\displaystyle a, b \ > 0$ and $\displaystyle a^6 = b^6 \ \Rightarrow \ \ a = b$

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.
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October 28th, 2014, 04:08 AM   #9
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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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