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# HELP PLZZZZ

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

THANK YOU!!

Guest May 29, 2017
#1
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\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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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
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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....!!!!

CPhill  May 29, 2017
#4
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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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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
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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  ???

CPhill  May 30, 2017
#7
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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
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How did you find the reciprocal  of   [ 1 + 3^(1/3) + 3^(2/3) ]   ????

CPhill  May 30, 2017
#9
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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
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Simplify $\frac{2\sqrt[3]9}{1 + \sqrt[3]3 + \sqrt[3]9}.$

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

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

0.919916176948 !

asinus  May 30, 2017
edited by asinus  May 30, 2017
edited by asinus  May 30, 2017
#11
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$$\dfrac{2\sqrt[3]{9}}{1+\sqrt[3]3+\sqrt[3]9}\\ \boxed{\text{Remember that }a^3 - b^3 = (a - b)(a^2+ab+b^2) }\\ =\dfrac{2\sqrt[3]9}{1+\sqrt[3]3+(\sqrt[3]{3})^2}\\ =\dfrac{(2\sqrt[3]9)(1-\sqrt[3]{3})}{(1-\sqrt[3]{3})(1+\sqrt[3]3+(\sqrt[3]{3})^2)}\\ =\dfrac{(2\sqrt[3]9)(1-\sqrt[3]{3})}{1^3-(\sqrt[3]{3})^3}\\ =\dfrac{(2\sqrt[3]9)(\sqrt[3]{3}-1)}{2}\\ =\sqrt[3]{9}\times(\sqrt[3]{3}-1)\\ =\sqrt[3]{27}-\sqrt[3]{9}\\ =3 - \sqrt[3]{9}\\ =3 - 3^{2/3}$$

Exactly what CPhill said!! :D

MaxWong  May 30, 2017
#12
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MAX: JUST SHEER BRILLIANCE !! CONGRATS.

Guest May 30, 2017