Sum This!

This is a classic telescoping problem. Using the difference of cubes formula:

$\frac{1}{\sqrt[3]{n^2} + \sqrt[3]{n(n+1)}+\sqrt[3]{(n+1)^2} }= \sqrt[3]{n+1} - \sqrt[3]{n}$

Therefore the given expression is:

$-1+\sqrt[3]{2}-\sqrt[3]{2}+\sqrt[3]{3}-\sqrt[3]{3}+\sqrt[3]{4} = \sqrt[3]{4}-1$ $\implies a = 4 \text{ and } b = -1 \implies a+b = \boxed{3}$

Notice that $a^3-b^3 = (a-b)(a^2+ab+b^2)$ Now look at each of the three denominators. For the first expression put $a = \sqrt[3]{2} , b = \sqrt[3]{1}$ to and multiply both the numerators and denominators by $a-b$ and then simplify to show that it is equal to $\sqrt[3]{2} - \sqrt[3]{1}$ . For the second expression use $a = \sqrt[3]{3} , b = \sqrt[3]{2}$ and for the third use $a = \sqrt[3]{4} , b = \sqrt[3]{3}$ to get the whole expression as equivalent to $\sqrt[3]{2} - \sqrt[3]{1} + \sqrt[3]{3} - \sqrt[3]{2} + \sqrt[3]{4} - \sqrt[3]{3}$ which simplifies to $\sqrt[3]{4} - 1$ so the answer is $4-1 = 3$

We can write 1/(a^2+ a.b + b^2 ) as (a-b)/(a^3-b^3) ..Complete expression boils down to (4)^(1/3) - 1 so a=4,b=-1 hence a+b is 3

Note that each of the denominator in this expression is a geometric series, and the first, second, third term respectively has common ration $\sqrt[3]{2}, \sqrt[3]{\frac{3}{2}}, \sqrt[3]{\frac{4}{3}}$ . Therefore the original expression becomes $\frac{\sqrt[3]{2}-1 }{\sqrt[3]{1}\left( 2-1 \right) } + \frac{\sqrt[3]{\frac{3}{2}}-1 }{\sqrt[3]{4}\left( \frac{3}{2}-1 \right) } + \frac{\sqrt[3]{\frac{4}{3}}-1 }{\sqrt[3]{9}\left( \frac{4}{3}-1 \right) } = \sqrt[3]{2}-\sqrt[3]{1}+\sqrt[3]{3}-\sqrt[3]{2}+\sqrt[3]{4}-\sqrt[3]{3} = \sqrt[3]{4}-1.$ So the answer is $\boxed{3}$ .