[SOLVED] Simplifying Radicals

Show using Bombelli's Method that $\displaystyle \sqrt[3]{\sqrt{325}+18} - \sqrt[3]{\sqrt{325}-18} = 3$

So I assume the form of the expressions by letting

$\displaystyle 1. \sqrt[3]{\sqrt{325}+18} = \sqrt{a} + b$

$\displaystyle 2. \sqrt[3]{\sqrt{325}-18} = \sqrt{a} - b$

Multiply 1 and 2 and you get: $\displaystyle 1 = a - b^2$

Cube both sides of 1 and you get: $\displaystyle a^{\frac{3}{2}} + 3ab + 3\sqrt{a}b + b^3 = 18 + \sqrt{325}$

Which means $\displaystyle a^{\frac{3}{2}} + 3\sqrt{a}{b} = \sqrt{325}$

and $\displaystyle 3ab + b^3 = 18$

But I can't get an answer from the equations. Any help would be greatly appreciated