Let $ n + 1 = m$ then we need to show that \[ \lfloor \sqrt [3]{m - 1} + \sqrt [3]{m} + \sqrt [3]{m + 1}\rfloor = \lfloor \sqrt [3]{27m - 1}\rfloor. \] For $ m < 10$ this equality can be verified instantly. For the rest of the proof assume that $ m\geq 10$. It easy to see (comparing derivatives) that for $ m\geq 10$, \[ (\sqrt [3]{m - 1} + \sqrt [3]{m + 1})^3 > 8m - \frac {8}{27}. \] Also, according to the power mean inequality, comparing the arithmetic and the power $ \frac{1}{3}$ means of $ m-1$ and $ m+1$, we have \[ 2\sqrt [3]{m} > \sqrt [3]{m - 1} + \sqrt [3]{m + 1}. \] Therefore, \[ 2\sqrt [3]{m} > \sqrt [3]{m - 1} + \sqrt [3]{m + 1} > 2\sqrt [3]{m - \frac {1}{27}} \] implying that \[ \sqrt [3]{27m} = 3\sqrt [3]{m} > \sqrt [3]{m - 1} + \sqrt [3]{m} + \sqrt [3]{m + 1} > 3\sqrt [3]{m - \frac {1}{27}} = \sqrt [3]{27m - 1}. \] Since there cannot be any integer strictly in between $ \sqrt [3]{27m - 1}$ and $ \sqrt [3]{27m}$, we have \[ \lfloor \sqrt [3]{m - 1} + \sqrt [3]{m} + \sqrt [3]{m + 1}\rfloor = \lfloor \sqrt [3]{27m - 1}\rfloor. \] Q.E.D.