\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.
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