세제곱근 합 계산

$\displaystyle \left( - 2 ^{\frac{4}{3}} + 250 ^{\frac{1}{3}} \right) ^{3} = \left( - 2 \times 2 ^{\frac{1}{3}} + 5 \times 2 ^{\frac{1}{3}} \right) ^{3} = \left( 3 \times 2 ^{\frac{1}{3}} \right) ^{3}$$= 27 \times 2 = 54$ [다른풀이] $\displaystyle \sqrt[3]{16} = a , \sqrt[3]{250} = b$라 두면 $\displaystyle \begin{aligned} \left( \sqrt[3]{- 16} + \sqrt[3]{250} \right) ^{3} = ( b - a ) ^{3} \\ = b ^{3} - 3 b ^{2} a + 3 ba ^{2} - a ^{3} \\ = 250 - 3 \sqrt[3]{250 \cdot 250 \cdot 16} + 3 \sqrt[3]{250 \cdot 16 \cdot 16} - 16 \\ = 250 - 3 \cdot \sqrt[3]{( 1000 ) ^{2}} + 3 \cdot \sqrt[3]{1000 \cdot 64} - 16 \\ = 250 - 3 \cdot 100 + 3 \cdot 40 - 10 = 54 \end{aligned}$