Points on the elliptic curve for Ramanujan-type cubic identities Given the rational Diophantine equation,
$$t^3 - t^2 - \tfrac{1}{3}(n^2 + n)t - \tfrac{1}{27}n^3=w^3\tag1$$
Two points are,
$$t_0 = 0\tag2$$
$$t_2 = \frac{-(1 + 2 n) (1 + 11 n + 42 n^2 + 14 n^3 + 13 n^4)}{9 (7 + 14 n + 24 n^2 + 17 n^3 + 19 n^4)}\tag3$$

Question: How do we find an intermediate point $\color{blue}{t_1}$ between $t_0$ and $t_2$; that is, one with numerator and denominator less than a quartic?

It's the missing piece in this family,
$$\sqrt[3]{t_0+x_1}+\sqrt[3]{t_0+x_2}+\sqrt[3]{t_0+x_3}= \sqrt[3]{z_0}\tag4$$
$$\sqrt[3]{\color{blue}{t_1}+x_1}+\sqrt[3]{\color{blue}{t_1}+x_2}+\sqrt[3]{\color{blue}{t_1}+x_3}= \sqrt[3]{z_1}\tag5$$
$$\sqrt[3]{t_2+x_1}+\sqrt[3]{t_2+x_2}+\sqrt[3]{t_2+x_3}= \sqrt[3]{z_2}\tag6$$
where,
$$z_0=-(2n+1)+3\sqrt[3]{\tfrac{n(n^2+n+1)}{3}}\tag7$$
$$z_2=\frac{-3(2 + n)^3  (1 + n + n^2)}{7 + 14 n + 24 n^2 + 17 n^3 + 19 n^4}\tag8$$
and the $x_i$ are the three roots of,
$$x^3 + x^2 - \tfrac{1}{3}(n^2 + n)x + \tfrac{1}{27}n^3=0\tag9$$
P.S. Once $\color{blue}{t_1}$ is found, and assuming it also has additional properties, it is easy to find $z_1$ as a rational root of a nonic. (The relations found by davidoff303 was just the special case $n=-3$.) There are infinitely many rational points $t_i$ but I am interested in those with the smallest height. 
 A: The cubic
\begin{equation*}
u^3-u^2-(n^2+n)u/3-n^3/27-w^3=0
\end{equation*}
can be shown to be equivalent to the elliptic curve
\begin{equation}
y^2=x^3+1296n^2(n^2+n+1)^2
\end{equation}
by using Nagell's algorithm and a computer algebra package. It does not have to be state-of-the-art software since I used an ancient MS-DOS version of Derive. The curve has torsion points of order $3$ when $x=0$.
The transformations are straightforward but lengthy, so are not given.
Tests suggest that the rank is greater than $0$ unless $n=1$. They suggest the rank is often greater than $1$.
Using simple experimental results it is possible to show that
\begin{equation}
(-4(n-1)(2n+1), \, 4(n+2)(7n^2+n+1))
\end{equation}
gives a simple non-torsion point on the curve, unless $n=1$.
Applying the transformations to this point and adding the torsion points just gives the values of $t$ not wanted, the quartic one given in the original question and two very similar points
\begin{equation*}
t=\frac{n^2(19n^4+17n^3+24n^2+14n+7)}{(1-n)(31n^4+41n^3+15n^2-7n+1)}
\end{equation*}
and
\begin{equation*}
t=\frac{n(n-1)(31n^4+41n^3+15n^2-7n+1)}{3(2n+1)(13n^4+14n^3+42n^2+11n+1)}
\end{equation*}
Doubling the point above gives a much more complex form. There might be massive cancellations reducing the corresponding value of $t$ but I did not try it.
