<s>[INST] Rationalize denominator: $\frac{1}{\sqrt[3]{5}+\sqrt[3]{1}+\sqrt[3]{6}}$ $$\frac{1}{\sqrt[3]{5}+\sqrt[3]{1}+\sqrt[3]{6}}$$
So this is what I thought: the square root of 1 is obviously one, so I have $1^3 +(\sqrt[3]{5} + \sqrt[3]{6})$. In my head I see that this is the first part for the sum of cubes formula. I multiplied with the rest of the formula so I can get $1^3 +(\sqrt[3]{5} + \sqrt[3]{6})^3$ in the denominator. Now when I try to do this bracket I have a problem, what do I do with this? $$11+3\sqrt[3]{25 \cdot 6}+3\sqrt[3]{5 \cdot 36}$$
How do I get rid of the square roots?
 [/INST] Use the identity: $$\dfrac{1}{\sqrt[3]{x}+\sqrt[3]{y}+\sqrt[3]{z}} = \dfrac{\sqrt[3]{x^2}+\sqrt[3]{y^2}+\sqrt[3]{z^2}-\sqrt[3]{xy}-\sqrt[3]{yz}-\sqrt[3]{zx}}{x+y+z - 3\sqrt[3]{xyz}}$$
and then use the difference of cube formula. Pretty sure there is no easier way to do this. 
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