Evaluating $\frac{1}{\sqrt{4}+\sqrt{6}+\sqrt{9}}+\frac{1}{\sqrt{9}+\sqrt{12}+\sqrt{16}}+\frac{1}{\sqrt{16}+\sqrt{20}+\sqrt{25}}=?$ It's a question (not hw) I bumped into few years back. Couldn't make any real progress with. Maybe you can help?
$$\frac{1}{\sqrt{4}+\sqrt{6}+\sqrt{9}}+\frac{1}{\sqrt{9}+\sqrt{12}+\sqrt{16}}+\frac{1}{\sqrt{16}+\sqrt{20}+\sqrt{25}}=?$$
Thanks.

The case of cube roots is probably more interesting than square roots; namely, simplifying
$$\frac{1}{\sqrt[3]{4}+\sqrt[3]{6}+\sqrt[3]{9}}+\frac{1}{\sqrt[3]{9}+\sqrt[3]{12‌​}+\sqrt[3]{16}}+\frac{1}{\sqrt[3]{16}+\sqrt[3]{20}+\sqrt[3]{25}}. \tag{$\circ$}$$
To evaluate this, as sos440 did in the comments, one notes the structure of the denominators are apparent as $a^2+ab+b^2$, which appears in the factorization in a difference of cubes, $b^3-a^3$. More generally, this sort of homogeneous polynomial results from using the geometric series formula on the common ratio $b/a$, but I digress. It is clear that we have
$$\frac{1}{a^2+ab+b^2}=\frac{b~-~a}{b^3-a^3} \tag{$\bullet$}$$
as the individual terms in $(\circ)$, with $b=\sqrt[3]{k+1}$ and $a=\sqrt[3]{k}$, for $k=2,3,4$. The denominators will all simply be $1$, and the three terms (the numerators) subsequently telescope:
$$\big(\sqrt[3]3-\sqrt[3]2\big)+\big(\sqrt[3]4-\sqrt[3]3\big)+\big(\sqrt[3]5-\sqrt[3]4\big)=\sqrt[3]5-\sqrt[3]2. \tag{$\square$}$$
