Calculate the radicals

Question

I want to calculate this expresion

$ \sqrt {1-2 \sqrt[3]{6} + \sqrt[3]{36}} $

And this expresion I want to simplify it

$ \frac {3 \sqrt{a}}{a} + a^{\frac{1}{6}} \sqrt[3]{a} - \frac{a^{\frac{2}{7}}}{\sqrt{a}} - \frac{3a^0}{\sqrt{a}} $ with a>0

Answer 1

$\sqrt[3]{36}=\sqrt[3]{6^2}=(\sqrt[3]6)^2$

So, $1-2\sqrt[3]6+\sqrt[3]{36}=1^2-2\cdot1\cdot\sqrt[3]6+(\sqrt[3]6)^2=?$

Answer 2

Hint:

The coefficients $1,-3,3,-1$ seem to indicate a perfect cube $(a^\alpha-a^\beta)^3$. If this is true, you will find terms $a^{3\alpha}$ and $-a^{3\beta}$.