HELP PLZZZZ

\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.

 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!!

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....!!!!

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) !!

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  ???

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.

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

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.

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 !

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

Exactly what CPhill said!! :D

MAX: JUST SHEER BRILLIANCE !! CONGRATS.