Let x = $$\root 6 \of {27} - \sqrt {6\frac{3}{4}} $$   and y = $$\frac{{\sqrt {45} + \sqrt {605} + \sqrt {245} }}{{\sqrt {80} + \sqrt {125} }},$$    then the value of x² + y² is:

Solution (By Examveda Team)

$$\eqalign{ & x = \root 6 \of {27} - \sqrt {6\frac{3}{4}} \cr & {x^2} = {\left( {{{27}^{\frac{1}{6}}} - \frac{{{{27}^{\frac{1}{2}}}}}{2}} \right)^2} \cr & {x^2} = {27^{\frac{2}{6}}} + \frac{{27}}{4} - \frac{{2 \times {{27}^{\frac{1}{6}}} \times {{27}^{\frac{1}{2}}}}}{2} \cr & {x^2} = 3 + \frac{{27}}{4} - 9 \cr & {x^2} = \frac{3}{4} \cr & y = \frac{{\sqrt {45} + \sqrt {605} + \sqrt {245} }}{{\sqrt {80} + \sqrt {125} }} \cr & y = \frac{{3\sqrt 5 + 11\sqrt 5 + 7\sqrt 5 }}{{4\sqrt 4 + 5\sqrt 5 }} \cr & y = \frac{7}{3} \cr & {y^2} = \frac{{49}}{9} \cr & {x^2} + {y^2} = \frac{3}{4} + \frac{{49}}{9} = \boxed{\frac{{223}}{{36}}} \cr} $$