세제곱근 합의 값

$\displaystyle ( a \pm \sqrt{b} ) ^{3}$$\displaystyle = ( a ^{3} + 3 ab ) \pm \sqrt{b} ( 3 a ^{2} + b )$ $a = 1$, $b = 3$일 때, $a ^{3} + 3 ab = 10$, $3 a ^{2} + b = 6$을 만족시킨다. ∴ $\displaystyle 10 + 2 \sqrt{27} = 10 + 6 \sqrt{3} = ( 1 + \sqrt{3} ) ^{3}$ $\displaystyle 10 - 2 \sqrt{27} = 10 - 6 \sqrt{3} = ( 1 - \sqrt{3} ) ^{3}$ ∴ $\displaystyle \sqrt[3]{10 + 2 \sqrt{27}} + \sqrt[3]{10 - 2 \sqrt{27}}$ $\displaystyle = \sqrt[3]{( 1 + \sqrt{3} ) ^{3}}$$\displaystyle + \sqrt[3]{( 1 - \sqrt{3} ) ^{3}}$ $\displaystyle = ( 1 + \sqrt{3} ) + ( 1 - \sqrt{3} )$ $= 2$