If $x=\frac12(\sqrt[3]{2009}-\frac{1}{\sqrt[3]{2009}})$, what is the value of $(x+\sqrt{1+x^2})^3$?

Question

If $x=\frac12(\sqrt[3]{2009}-\frac{1}{\sqrt[3]{2009}})$, what is the value of $(x+\sqrt{1+x^2})^3$?

I solved this problem as follow,

Assuming $\sqrt[3]{2009}=\alpha$ , we have $x=\frac12(\alpha-\frac1{\alpha})$ and $$(x+\sqrt{1+x^2})^3=\left[\frac12(\alpha-\frac1{\alpha}) +\sqrt{1+\frac14(\alpha^2+\frac1{\alpha^2}-2)}\right]^3=\left[(\frac{\alpha}2-\frac1{2\alpha}) +\sqrt{\frac{\alpha^2}4+\frac1{4\alpha^2}+\frac12)}\right]^3=\left(\frac{\alpha}2-\frac1{2\alpha}+\left|\frac{\alpha}2+\frac1{2\alpha}\right|\right)^3=\alpha^3=2009$$ I'm wondering, is it possible to solve this problem with other approaches?

Answer 1

Assuming $\sqrt[3]{2009}=\alpha$ , we have $x=\dfrac12\left(\alpha-\dfrac1{\alpha}\right)$

We also have:

$$ \begin{align} x = \frac{1}{2}\left(x+\sqrt{1+x^2} \;+\; x-\sqrt{1+x^2}\right) = \frac{1}{2}\left(x+\sqrt{1+x^2} \;-\; \frac{1}{x+\sqrt{1+x^2}}\right) \end{align} $$

Comparing with $\,x = \dfrac12\left(\alpha-\dfrac1{\alpha}\right)\,$ it follows that $\,x+\sqrt{1+x^2} = \alpha\,$, because the function $\,f(t) = \dfrac{1}{2}\left(t - \dfrac{1}{t}\right)\,$ is injective on $\,\mathbb R^+\,$.

Answer 2

Here's a rather elegant approach that involves pattern matching. Let $x=\sinh t$. Then

$$(x+\sqrt{1+x^2})^3 = (\sinh t + \cosh t)^3 = e^{3t}$$

And from comparing definitions

$$\sinh t = \frac{1}{2}(e^t-e^{-t})$$

$$x = \frac{1}{2}\left(\sqrt[3]{2009}-\frac{1}{\sqrt[3]{2009}}\right)$$

clearly, $e^t=\sqrt[3]{2009}$, thus

$$e^{3t} = 2009$$