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Simplify  and rationalize the denominator. The result can be expressed in the form , where  and  are integers. What is the value of the sum ?

Guest Jan 13, 2018

#1
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$$\quad\,\frac{3}{\sqrt[5]{16}}\,+\,\frac{1}{\sqrt3} \\~\\ =\,\frac{3}{\sqrt[5]{2^4}}\,+\,\frac{1}{\sqrt3} \\~\\ =\,\frac{3}{2^\frac45}\,+\,\frac1{3^\frac12}\\~\\ =\,\frac{3\,\cdot\,2^\frac15}{2^\frac45\,\cdot\,2^\frac15}\,+\,\frac{1\,\cdot\,3^\frac12}{3^\frac12\,\cdot\,3^\frac12}\\~\\ =\,\frac{3\,\cdot\,2^\frac15}{2}\,+\,\frac{3^\frac12}{3}\\~\\ =\,\frac{3\,\cdot\,2^\frac15\,\cdot\,3}{2\,\cdot\,3}\,+\,\frac{3^\frac12\,\cdot\,2}{3\,\cdot\,2}\\~\\ =\,\frac{3^2\,\cdot\,2^\frac15}{3\cdot2}\,+\,\frac{3^\frac12\,\cdot\,2}{3\cdot2}\\~\\ =\,\frac{3^2\,\cdot\,2^\frac15\,+\,3^\frac12\,\cdot\,2}{3\cdot2}\\~\\ =\,\frac{3^2\sqrt[5]{2}\,+\,2\sqrt3}{3\cdot2}$$

a = 3    and    b = 2     so     a + b  =  3 + 2  =  5

hectictar  Jan 13, 2018
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#1
+5924
+1
$$\quad\,\frac{3}{\sqrt[5]{16}}\,+\,\frac{1}{\sqrt3} \\~\\ =\,\frac{3}{\sqrt[5]{2^4}}\,+\,\frac{1}{\sqrt3} \\~\\ =\,\frac{3}{2^\frac45}\,+\,\frac1{3^\frac12}\\~\\ =\,\frac{3\,\cdot\,2^\frac15}{2^\frac45\,\cdot\,2^\frac15}\,+\,\frac{1\,\cdot\,3^\frac12}{3^\frac12\,\cdot\,3^\frac12}\\~\\ =\,\frac{3\,\cdot\,2^\frac15}{2}\,+\,\frac{3^\frac12}{3}\\~\\ =\,\frac{3\,\cdot\,2^\frac15\,\cdot\,3}{2\,\cdot\,3}\,+\,\frac{3^\frac12\,\cdot\,2}{3\,\cdot\,2}\\~\\ =\,\frac{3^2\,\cdot\,2^\frac15}{3\cdot2}\,+\,\frac{3^\frac12\,\cdot\,2}{3\cdot2}\\~\\ =\,\frac{3^2\,\cdot\,2^\frac15\,+\,3^\frac12\,\cdot\,2}{3\cdot2}\\~\\ =\,\frac{3^2\sqrt[5]{2}\,+\,2\sqrt3}{3\cdot2}$$