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Simplify $\frac{2\sqrt[3]9}{1 + \sqrt[3]3 + \sqrt[3]9}.$

THANK YOU!!

Guest May 29, 2017
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12+0 Answers

 #1
avatar+89843 
0

\frac{2\sqrt[3]9}{1 + \sqrt[3]3 + \sqrt[3]9}.

 

\(\frac{2\sqrt[3]9}{1 + \sqrt[3]3 + \sqrt[3]9}\\\)

 

I do not think this can be simplified.

Melody  May 29, 2017
 #2
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+1

 simplify | (2 9^(1/3))/(1 + 3^(1/3) + 9^(1/3)) =3 - 3^(2/3)*

 

* My Mathematica 11 Home Package came up with that answer but will not give details or steps as to how they got that answer!!

Guest May 29, 2017
 #3
avatar+75376 
+1

2*9^(1/3)  / [ 1 + 3^(1/3) + 9^(1/3) ]  =

 

2*3^(2/3) /  [ 1 + 3^(1/3) + 3^(2/3) ] 

 

WolframAlpha shows that this can actually be simplified to  

 

3  - 3^(2/3)

 

How they got there, I don't know...I tried to use a conjugate, but didn't get anywhere.....maybe I didn't follow it through far enough???....maybe some other mathematician knows how to simplify this step-by-step....!!!!

 

 

cool cool cool

CPhill  May 29, 2017
 #4
avatar+89843 
+1

Yes I saw that too Chris and I also have do not know how they did that. 

Melody  May 30, 2017
 #5
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+2

I got it! =2 3^(2/3) (3^(1/3)/2 - 1/2)=2[3/2 - 3^(2/3) /2]=[6/2 - 3^(2/3) =3 - 3^(2/3) !!

Guest May 30, 2017
 #6
avatar+75376 
0

 

Yep, guest......that looks to be correct....good job  !!!

 

BTW -  the denominator  is   ( [ 1 + 3^(1/3) + 3^(2/3)] [ 3^(1/3)/2 - 1/2])  which simplifies to 1

 

What tipped you off  ???

 

 

 

cool cool cool

CPhill  May 30, 2017
 #7
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0

I took the reciprocal of the denominator for which I got: 1/2 (3^(1/3) - 1) =3^(1/3)/2 - 1/2. Lastly, I multiplied this by the first term which gave me the result above.

Guest May 30, 2017
 #8
avatar+75376 
0

How did you find the reciprocal  of   [ 1 + 3^(1/3) + 3^(2/3) ]   ????

 

 

cool cool cool

CPhill  May 30, 2017
 #9
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0

Thanks to my Mathematica 11 package !! By the way, using W/A also gives the same result ! Of course, W/A gets its results from Mathematica.

Guest May 30, 2017
 #10
avatar+6722 
0

Simplify $\frac{2\sqrt[3]9}{1 + \sqrt[3]3 + \sqrt[3]9}.$

 

\(\frac{2\sqrt[3]9}{1 + \sqrt[3]3 + \sqrt[3]9}\\Perhaps \ the \ reciprocal?\\\color{blue}\frac{{1 + \sqrt[3]3 + \sqrt[3]9}}{2\sqrt[3]{9}}= \frac{1}{2\sqrt[3]{9}}+\frac{1}{2\sqrt[3]{3}}+\frac{1}{2}\)

 

\(\frac{2\sqrt[3]9}{1 + \sqrt[3]3 + \sqrt[3]9}=1/( \frac{1}{2\sqrt[3]{9}}+\frac{1}{2\sqrt[3]{3}}+\frac{1}{2})\)

 

laugh 0.919916176948 !

asinus  May 30, 2017
edited by asinus  May 30, 2017
edited by asinus  May 30, 2017
 #11
avatar+6765 
+1

\(\dfrac{2\sqrt[3]{9}}{1+\sqrt[3]3+\sqrt[3]9}\\ \boxed{\text{Remember that }a^3 - b^3 = (a - b)(a^2+ab+b^2) }\\ =\dfrac{2\sqrt[3]9}{1+\sqrt[3]3+(\sqrt[3]{3})^2}\\ =\dfrac{(2\sqrt[3]9)(1-\sqrt[3]{3})}{(1-\sqrt[3]{3})(1+\sqrt[3]3+(\sqrt[3]{3})^2)}\\ =\dfrac{(2\sqrt[3]9)(1-\sqrt[3]{3})}{1^3-(\sqrt[3]{3})^3}\\ =\dfrac{(2\sqrt[3]9)(\sqrt[3]{3}-1)}{2}\\ =\sqrt[3]{9}\times(\sqrt[3]{3}-1)\\ =\sqrt[3]{27}-\sqrt[3]{9}\\ =3 - \sqrt[3]{9}\\ =3 - 3^{2/3}\)

Exactly what CPhill said!! :D

MaxWong  May 30, 2017
 #12
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0

MAX: JUST SHEER BRILLIANCE !! CONGRATS.

Guest May 30, 2017

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