Which positive integer \(n\) satisfies the equation \[\frac{3}{n^3}+\frac{4}{n^3}+\frac{5}{n^3}+\dotsb +\frac{n^3-5}{n^3}+\frac{n^3-4}{n^3}+\frac{n^3-3}{n^3}=60?\]

This is an arithmetic progression with common difference \(\dfrac{1}{n^3}\). There are \(n^3-5\) terms.

The formula for the sum of \(k\) terms of an arithmetic progression \(a_1 + \dotsb + a_k\) is given by \(\dfrac{k}{2}\left(a_1 + a_k\right).\)

Alternatively, if the first term is \(a\), and the common difference is \(d\), and if there are \(n\) terms, then the sum of the first \(n\) terms is \(S_n = \dfrac{n}{2}(2a + (n-1)d)\).

Their sum is therefore \[60=\frac{n^3-5}{2}\left(\frac{3}{n^3}+\frac{n^3-3}{n^3}\right)=\frac{n^3-5}{2}\left(\frac{n^3}{n^3}\right)=\frac{n^3-5}{2},\] so we can rearrange to get \[n^3-5=120\implies n^3=125\implies n=5.\]

Alternatively, we could have written the sum as

\[\frac{3}{n^3}+\frac{4}{n^3}+\frac{5}{n^3}+\dotsb +\frac{n^3-5}{n^3}+\frac{n^3-4}{n^3}+\frac{n^3-3}{n^3}=\frac{3+4+\dotsb+(n^3-3)}{n^3}\]

Now we can treat the numerator as an arithmetic series giving \[\frac{\frac{(n^3-3)(n^3-2)}{2}-1-2}{n^3} =\frac{n^6-5 n^3+6-6}{2 n^3} =\frac{n^3-5}{2} = 60, n = 5.\]