MHB Which Expression is Larger in This Mathematical Comparison?

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The mathematical comparison reveals that the expressions $(2^{\frac{1}{3}}-1)^{\frac{1}{3}}$ and $\left( \dfrac{1}{9} \right)^{\frac{1}{3}}-\left( \dfrac{2}{9} \right)^{\frac{1}{3}}+\left( \dfrac{4}{9} \right)^{\frac{1}{3}}$ are equal. By letting $\alpha = 2^{1/3}$ and manipulating the equations, it is shown that both expressions simplify to the same value. The calculations involve recognizing geometric series and applying the sum formula. Ultimately, the conclusion is that both expressions yield the same result, confirming their equality. This illustrates a successful resolution of the mathematical comparison.
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Which is larger?

$(2^{\frac{1}{3}}-1)^{\frac{1}{3}}$ or $\left( \dfrac{1}{9} \right)^{\frac{1}{3}}-\left( \dfrac{2}{9} \right)^{\frac{1}{3}}+\left( \dfrac{4}{9} \right)^{\frac{1}{3}}$
 
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[sp]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.[/sp]
 
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 $(2^{\frac{1}{3}}-1)^{\frac{1}{3}}=\left( \dfrac{1}{9} \right)^{\frac{1}{3}}-\left( \dfrac{2}{9} \right)^{\frac{1}{3}}+\left( \dfrac{4}{9} \right)^{\frac{1}{3}}$
 
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