# Which is larger?

#### anemone

##### MHB POTW Director
Staff member
Which is larger?

$(2^{\frac{1}{3}}-1)^{\frac{1}{3}}$ or $\left( \dfrac{1}{9} \right)^{\frac{1}{3}}-\left( \dfrac{2}{9} \right)^{\frac{1}{3}}+\left( \dfrac{4}{9} \right)^{\frac{1}{3}}$

#### Opalg

##### MHB Oldtimer
Staff member
Let $\alpha = 2^{1/3}$. Then \begin{aligned}\left( \left( \tfrac{1}{9} \right)^{1/3}-\left( \tfrac{2}{9} \right)^{1/3}+\left( \tfrac{4}{9} \right)^{1/3}\right)^{\!\!3} = \tfrac19(1-\alpha+\alpha^2)^3 &= \tfrac19(1-3\alpha +6\alpha^2 -7\alpha^3 +6\alpha^4 -3\alpha^5 +\alpha^6) \\ &= \tfrac19(1-3\alpha +6\alpha^2 -14 +12\alpha -6\alpha^2 + 4) = \tfrac19(-9+9\alpha) = -1+\alpha.\end{aligned} So $(2^{1/3}-1)^{1/3}$ and $\left( \tfrac{1}{9} \right)^{1/3}-\left( \tfrac{2}{9} \right)^{1/3}+\left( \tfrac{4}{9} \right)^{1/3}$ are equal.

#### anemone

##### MHB POTW Director
Staff member
Thanks for participating, Opalg! I often get the feeling that you solve all the challenging problems like a breeze and I admire your ability and capability very much!

My solution:

I first let

$y=(2^{\frac{1}{3}}-1)^{\frac{1}{3}}$

and we can now undo cube root by cubing both sides of the equation and get

$y^3=2^{\frac{1}{3}}-1$

I then let $x=\left( \dfrac{1}{9} \right)^{\frac{1}{3}}-\left( \dfrac{2}{9} \right)^{\frac{1}{3}}+\left( \dfrac{4}{9} \right)^{\frac{1}{3}}$

Notice that $\left( \dfrac{1}{9} \right)^{\frac{1}{3}}-\left( \dfrac{2}{9} \right)^{\frac{1}{3}}+\left( \dfrac{4}{9} \right)^{\frac{1}{3}}$ is a geometric series with first term and common ratio of $\left(\dfrac{1}{9} \right)^{\frac{1}{3}}$ and $-(2)^{\frac{1}{3}}$ and $x$ is the sum of the first three terms of the series, hence, by using the sum of the first $n$ terms of a geometric series, we have

$x=\dfrac{\left( \dfrac{1}{9} \right)^{\frac{1}{3}}\left( 1-(-(2^{\frac{1}{3}})^3 \right)}{1-(-(2^{\frac{1}{3}})}=\dfrac{3^{\frac{1}{3}}}{1+2^{ \frac{1}{3}}}$

$\therefore x^3=\dfrac{3}{(1+2^{ \frac{1}{3}})^3}=\dfrac{3}{3+3(2^{ \frac{1}{3}}+2^{ \frac{2}{3}})}=\dfrac{1}{1+2^{ \frac{1}{3}}+2^{ \frac{2}{3}}}$

and notice again that $1+2^{ \frac{1}{3}}+2^{ \frac{2}{3}}$ is another geometric series with first term and common ratio of $1$ and $(2)^{\frac{1}{3}}$ hence

$x^3=\dfrac{1}{1+2^{ \frac{1}{3}}+2^{ \frac{2}{3}}}$

$\;\;\;\;=\dfrac{1}{\dfrac{1((2^{ \frac{1}{3}})^3-1)}{2^{ \frac{1}{3}}-1}}$

$\;\;\;\;=\dfrac{1}{\dfrac{1}{2^{ \frac{1}{3}}-1}}$

$\;\;\;\;=2^{ \frac{1}{3}}-1$

$\;\;\;\;=y^3$

This implies $x=y$.

Therefore, we can conclude now that $(2^{\frac{1}{3}}-1)^{\frac{1}{3}}=\left( \dfrac{1}{9} \right)^{\frac{1}{3}}-\left( \dfrac{2}{9} \right)^{\frac{1}{3}}+\left( \dfrac{4}{9} \right)^{\frac{1}{3}}$