이중근호 x³+y³

$\displaystyle \begin{aligned} x ^{2} & = 11 - 4 \sqrt{4 + 2 \sqrt{3}} = 11 - 4 ( \sqrt{3} + 1 ) = 7 - 2 \sqrt{12} \end{aligned}$ $\displaystyle \therefore x = \sqrt{7 - 2 \sqrt{12}} = 2 - \sqrt{3}$ $\displaystyle \begin{aligned} y ^{2} & = 11 + 4 \sqrt{4 - 2 \sqrt{3}} = 11 + 4 ( \sqrt{3} - 1 ) = 7 + 2 \sqrt{12} \end{aligned}$ $\displaystyle \therefore y = \sqrt{7 + 2 \sqrt{12}} = 2 + \sqrt{3}$ $x + y = 4 , xy = 1$이므로 $x ^{3} + y ^{3} = ( x + y ) ^{3} - 3 xy ( x + y ) = 52$