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The sum of the first $\large n$ natural numbers $\large 1 + 2 + 3 + 4 + \ ... \ + n$ is given by the formula

$\displaystyle \large \frac{n(n + 1)}{2}$

The sum of the first $\large n$ cubes $\large 1^3 + 2^3 + 3^3 + 4^3 + \ ... \ + n^3$ is the square of the sum of the first n natural numbers:

$\displaystyle \large {\bigg(\frac{n(n + 1)}{2}\bigg)}^2$

Thus, we can rewrite the expression as

$\displaystyle \large \left( \dfrac{1+2+3+4+\cdots+n} {1+8+27+64+\cdots+n^3} \right)^{-1}$

$\displaystyle \large = \Bigg[{\frac{\frac{n(n + 1)}{2}}{\Bigg({\frac{n(n + 1)}{2}\Bigg)}^2}}\Bigg] ^{- 1}$

$\displaystyle \large = {\Bigg[{\frac {1}{\frac {n(n + 1)}{2}}}\Bigg]}^{- 1}$

$\displaystyle \large = \frac{n(n + 1)}{2}$