<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] Since $$a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$$
$$\frac{1}{\sqrt[3]{5}+\sqrt[3]{1}+\sqrt[3]{6}}=\frac{\sqrt[3]{5^2}+1+\sqrt[3]{36} - \sqrt[3]{5}-\sqrt[3]{6}-\sqrt[3]{30}}{12-3\sqrt[3]{30}}$$
and since $a^3-b^3=(a-b)(a^2+ab+b^2)$$$\frac{1}{\sqrt[3]{30}-4}=\frac{\sqrt[3]{30^2}+4\sqrt[3]{30}+16}{30-64}$$
you just have to multiply them to obtain the answer, but do not forget $-\frac14$ factor.
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