Question

 The fraction 2(2^1/2 +6^1/2)/ 3(2^1/2 +3^1/4) is equal to ?

Answer 1

Multiply numerator and denom by  √2 - 3^1/4
=  2  (2  - 12^1/4 + √12 - 108^1/4 ) / 3(2 - √3)              now multiply by 2 + √3
=  2  (4  - 192^1/4 +  4√3  - 1728^1/4 + 2√3 + 6 - 9 √12 ) / 3
=  2 ( 10  - 12 √3 - 2 (√3+1)(12)^1/4 ) /3
 

Answer 2

$\frac{2(2^\frac{1}{2}+6^\frac{1}{2})}{3(2^\frac{1}{2}+3^\frac{1}{4})}=\frac{2(\sqrt2+\sqrt6)}{3(\sqrt2+\sqrt[4]3)}\cdot\frac{\sqrt2-\sqrt[4]3}{\sqrt2-\sqrt[4]3}=\frac{2(\sqrt2+\sqrt6)(\sqrt2-\sqrt[4]3)}{3[(\sqrt2)^2-(\sqrt[4]3)^2}\\\\=\frac{2\cdot2-2\sqrt2\cdot\sqrt[4]3+2\sqrt{12}-2\sqrt6\cdot\sqrt[4]3}{3(2-\sqrt3)}=\frac{4-2\sqrt[4]{4\cdot3}+2\sqrt{4\cdot3}-2\sqrt[4]{36\cdot3}}{3(2-\sqrt3)}$

$=\frac{4-\sqrt[4]{12}+4\sqrt3-2\sqrt[4]{108}}{3(2-\sqrt3)}\cdot\frac{2+\sqrt3}{2+\sqrt3}\\\\=\frac{8-2\sqrt[4]{12}+8\sqrt3-4\sqrt[4]{108}+4\sqrt3-\sqrt[4]{12\cdot9}+4\cdot3-2\sqrt[4]{108\cdot9}}{3[2^2-(\sqrt3)^2]}\\\\=\frac{8-2\sqrt[4]{12}+12\sqrt3-4\sqrt[4]{108}-\sqrt[4]{108}+12-2\sqrt[4]{81\cdot12}}{3(4-3)}\\\\=\frac{20-2\sqrt[4]{12}+12\sqrt3-5\sqrt[4]{108}-6\sqrt[4]{12}}{3}\\\\=\frac{20-8\sqrt[4]{12}+12\sqrt3-5\sqrt[4]{108}}{3}$