Simplify and rationalize the denominator. The result can be expressed in the form , where and are integers. What is the value of the sum ?
\(\quad\,\frac{3}{\sqrt[5]{16}}\,+\,\frac{1}{\sqrt3} \\~\\ =\,\frac{3}{\sqrt[5]{2^4}}\,+\,\frac{1}{\sqrt3} \\~\\ =\,\frac{3}{2^\frac45}\,+\,\frac1{3^\frac12}\\~\\ =\,\frac{3\,\cdot\,2^\frac15}{2^\frac45\,\cdot\,2^\frac15}\,+\,\frac{1\,\cdot\,3^\frac12}{3^\frac12\,\cdot\,3^\frac12}\\~\\ =\,\frac{3\,\cdot\,2^\frac15}{2}\,+\,\frac{3^\frac12}{3}\\~\\ =\,\frac{3\,\cdot\,2^\frac15\,\cdot\,3}{2\,\cdot\,3}\,+\,\frac{3^\frac12\,\cdot\,2}{3\,\cdot\,2}\\~\\ =\,\frac{3^2\,\cdot\,2^\frac15}{3\cdot2}\,+\,\frac{3^\frac12\,\cdot\,2}{3\cdot2}\\~\\ =\,\frac{3^2\,\cdot\,2^\frac15\,+\,3^\frac12\,\cdot\,2}{3\cdot2}\\~\\ =\,\frac{3^2\sqrt[5]{2}\,+\,2\sqrt3}{3\cdot2}\)
a = 3 and b = 2 so a + b = 3 + 2 = 5
\(\quad\,\frac{3}{\sqrt[5]{16}}\,+\,\frac{1}{\sqrt3} \\~\\ =\,\frac{3}{\sqrt[5]{2^4}}\,+\,\frac{1}{\sqrt3} \\~\\ =\,\frac{3}{2^\frac45}\,+\,\frac1{3^\frac12}\\~\\ =\,\frac{3\,\cdot\,2^\frac15}{2^\frac45\,\cdot\,2^\frac15}\,+\,\frac{1\,\cdot\,3^\frac12}{3^\frac12\,\cdot\,3^\frac12}\\~\\ =\,\frac{3\,\cdot\,2^\frac15}{2}\,+\,\frac{3^\frac12}{3}\\~\\ =\,\frac{3\,\cdot\,2^\frac15\,\cdot\,3}{2\,\cdot\,3}\,+\,\frac{3^\frac12\,\cdot\,2}{3\,\cdot\,2}\\~\\ =\,\frac{3^2\,\cdot\,2^\frac15}{3\cdot2}\,+\,\frac{3^\frac12\,\cdot\,2}{3\cdot2}\\~\\ =\,\frac{3^2\,\cdot\,2^\frac15\,+\,3^\frac12\,\cdot\,2}{3\cdot2}\\~\\ =\,\frac{3^2\sqrt[5]{2}\,+\,2\sqrt3}{3\cdot2}\)
a = 3 and b = 2 so a + b = 3 + 2 = 5