Show $\frac{2}{\sqrt[3]2}-\frac{1}{2(\sqrt[3]2-1)}+\left(\frac{9}{2\sqrt[3]4}-\frac{9}{4}\right)^{\frac{1}{3}}=\frac{1}{2}$? Prove that:
$$\frac{2}{\sqrt[3]2}-\frac{1}{2(\sqrt[3]2-1)}+\left(\frac{9}{2\sqrt[3]4}-\frac{9}{4}\right)^{\frac{1}{3}}=\frac{1}{2}$$

The LHS is irrational number and RHS is rational number. May be this is almost a $\frac{1}{2}$. 
According to the calculator it is a $\frac{1}{2}$
I tried to cube both sides but it looked too messy.
 A: Let me try. 
Note that $$(1+\sqrt[3]{2})^3 = 3(1+\sqrt[3]{2}+\sqrt[3]{4}).$$
Now we have:
$$LHS = \sqrt[3]{4} - \frac{1}{2}(\sqrt[3]{4} + \sqrt[3]{2} + 1) + \left(\frac{9}{4(\sqrt[3]{4}+\sqrt[3]{2}+1) }\right)^{\frac{1}{3}} = \sqrt[3]{4} - \frac{1}{2}(\sqrt[3]{4} + \sqrt[3]{2} + 1) + \left(\frac{27}{4(\sqrt[3]{2}+1)^3 }\right)^{\frac{1}{3}} = \sqrt[3]{4} - \frac{1}{2}(\sqrt[3]{4} + \sqrt[3]{2} + 1) + \frac{3}{\sqrt[3]{4}(\sqrt[3]{2}+1)} = \frac{1}{2}(\sqrt[3]{4} - \sqrt[3]{2} - 1) + \frac{\sqrt[3]{2}(\sqrt[3]{4}-\sqrt[3]{2}+1)}{2} = \frac{1}{2}.$$
A: Here's the proof I would use... note that I skip a few steps, specifically on simplifying the far right term on the LHS. Likewise, I apologize for any typos... my computer crashed thrice while writing this
$$\frac{2}{\sqrt[3]2}-\frac{1}{2(\sqrt[3]2-1)}+\left(\frac{9}{2\sqrt[3]4}-\frac{9}{4}\right)^{\frac{1}{3}}=\frac{1}{2}$$
$$= 2^{2/3} - \frac{1}{2(\sqrt[3]2-1)} + \left(\frac{9}{4}\left(2^{1/3} - 1\right)\right)^{1/3}$$
$$= 2^{2/3} - \frac{1}{2(\sqrt[3]2-1)} + 1+\frac{1}{2^{2/3}}-\frac{1}{2^{1/3}}$$
$$= \frac{1}{2}\left(2^{2/3}(2-1) - \frac{1}{\sqrt[3]2-1} + 2+2^{1/3}\right)$$
$$= \frac{1}{2}\left(2^{2/3} - \frac{1}{2^{1/3}-1} + 2+2^{1/3}\right)$$
$$= \frac{1}{2(\sqrt[3]2-1)}\left(2-2^{2/3} - 1 + 2^{4/3} - 2+2^{2/3} - 2^{1/3}\right)$$
$$= \frac{1}{2(2^{1/3}-1)}\left( 2^{1/3} - 1\right)$$
$$=\color{red}{\frac{1}{2}}$$  
Note here that the LHS is rational... the product of irrational numbers can trivially yield a rational number (take $\sqrt{2}\sqrt{2} = 2$) as can related operations
