# Rationalize denominator: $\frac{1}{\sqrt{5}+\sqrt{1}+\sqrt{6}}$

$$\frac{1}{\sqrt{5}+\sqrt{1}+\sqrt{6}}$$ So this is what I thought: the square root of 1 is obviously one, so I have $1^3 +(\sqrt{5} + \sqrt{6})$. 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 $1^3 +(\sqrt{5} + \sqrt{6})^3$ in the denominator. Now when I try to do this bracket I have a problem, what do I do with this? $$11+3\sqrt{25 \cdot 6}+3\sqrt{5 \cdot 36}$$

How do I get rid of the square roots?

## 2 Answers

Since $$a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$$

$$\frac{1}{\sqrt{5}+\sqrt{1}+\sqrt{6}}=\frac{\sqrt{5^2}+1+\sqrt{36} - \sqrt{5}-\sqrt{6}-\sqrt{30}}{12-3\sqrt{30}}$$

and since $a^3-b^3=(a-b)(a^2+ab+b^2)$$\frac{1}{\sqrt{30}-4}=\frac{\sqrt{30^2}+4\sqrt{30}+16}{30-64}$$ you just have to multiply them to obtain the answer, but do not forget$-\frac14$factor. • Yes, of course. This is iconic formula. – Hedgehog Oct 25 '17 at 21:07 Use the identity: $$\dfrac{1}{\sqrt{x}+\sqrt{y}+\sqrt{z}} = \dfrac{\sqrt{x^2}+\sqrt{y^2}+\sqrt{z^2}-\sqrt{xy}-\sqrt{yz}-\sqrt{zx}}{x+y+z - 3\sqrt{xyz}}$$ and then use the difference of cube formula. Pretty sure there is no easier way to do this. • I then have$3\sqrt{30}$. What am I supposed to do with that? – RiktasMath Oct 25 '17 at 21:29 • @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{30} + (3 \sqrt{30})^2\$. – Jacob Oct 25 '17 at 21:38