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$$\frac{1}{\sqrt[3]{5}+\sqrt[3]{1}+\sqrt[3]{6}}$$ 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?

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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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  • $\begingroup$ Yes, of course. This is iconic formula. $\endgroup$ – Hedgehog Oct 25 '17 at 21:07
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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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  • $\begingroup$ I then have $3\sqrt[3]{30}$. What am I supposed to do with that? $\endgroup$ – RiktasMath Oct 25 '17 at 21:29
  • $\begingroup$ @RiktasMath as he says in the answer, you need to use the difference of cubes formula. So you should multiply the top and bottom by $12^2+12\cdot 3 \sqrt[3]{30} + (3 \sqrt[3]{30})^2$. $\endgroup$ – Jacob Oct 25 '17 at 21:38

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