<s>[INST] If $\sqrt[3]x-\sqrt[3]{x-14}=2$, find $x - 1/x$ 
$$ \sqrt[3]x-\sqrt[3]{x-14}=2$$
  Then find $x - \frac{1}{x} = ?$ 

I've tried to take third power of both sides but I couldn't find anything. 
 [/INST] The most direct method:
$\sqrt[3]x-\sqrt[3]{x-14}=2$
$\implies\ \sqrt[3]{x-14}=\sqrt[3]x-2$
$\implies\ x-14=\left(\sqrt[3]x-2\right)^3=x-6(\sqrt[3]x)^2+12(\sqrt[3]x)-8$
$\implies\ 6y^2-12y-6=0$ where $y=\sqrt[3]x$
$\implies\ y=1\pm\sqrt2$
$\implies\ y^3=x=7\pm5\sqrt2$
$\implies\ \dfrac1x=\pm5\sqrt2-7$
Hence $x-\dfrac1x=14$
</s>