There are no positive integers $a,b,n$ such that $n^2 < a^3 < b^3 < (n + 1)^2$. Suppose for the sake of contradiction that there are some positive integers $a,b,n$ such that $n^2 < a^3 < b^3 < (n + 1)^2$.We have $b>a$,so $b\ge a+1$,which implies $n^2<a^3<(a+1)^3<(n+1)^2$.But $a>\sqrt[3]{n^2}$,so $(a+1)^3=a^3+3a^2+3a+1\ge n^2+3\sqrt[3]{n^4}+3\sqrt[3]{n^2}+1\ge n^2+2n+1=$ $(n+1)^2$,contradiction!This ends our proof.

It is easier, perhaps, to note that between any two cubes inclusive there is a square, since for any $a$ there is $x$ with $a\sqrt{a} \le x \le (a + 1)\sqrt{a + 1} = a\sqrt{a + 1} + \sqrt{a + 1}$, showing that the range for $x$ is much larger than $1$. But then $n^2 < a^3 \le x^2 \le b^3 < (n + 1)^2, a clear contradiction for this proves $n < x < n + 1$. This is perhaps a nice way to show how the square are more "compact" in all ways, and best of all it extends easily without weird binomial theorem ideas (although admittedly, if written correctly, require pretty much no effort).