$\sqrt[3]{\dfrac{1}{8}} + \sqrt[3]{\dfrac{1}{27}}$ equals

$\sqrt[3]{\dfrac{1}{8}} + \sqrt[3]{\dfrac{1}{27}}$ equals
  1. $\dfrac{5}{6}$
  2. $\dfrac{1}{6}$
  3. $\dfrac{2}{5}$
  4. $\dfrac{1}{12}$

Solution

$\dfrac{1}{2} + \dfrac{1}{3} = \dfrac{5}{6}$.

Asked in: IMO

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