Given the cubic, $$z^3-ez^2+fz-1=0$$ Let $z_1,z_2,z_3$ be the three roots. Define, $$x_n = z_1^{1/n}+ z_2^{1/n}+ z_3^{1/n}$$ $$y_n = (z_1z_2)^{1/n}+ (z_1z_3)^{1/n}+(z_2z_3)^{1/n}$$ Ramanujan found that, $$x_3^3-e = y_3^3-f= 3(x_3y_3-1) \tag{1}$$ $$ \frac{x_5^5-e}{x_5^2-y_5} = \frac{y_5^5-f}{y_5^2-x_5} = 5(x_5y_5-1) \tag{2}$$ $$ \frac{x_7^7-e}{(x_7^2-y_7)^2+x_7} = \frac{y_7^7-f}{(y_7^2-x_7)^2+y_7} = 7(x_7y_7-1) \tag{3}$$ He stops here, but is there a generalization to $p=11$, $$ \frac{x_{11}^{11}-e}{P_1} = \frac{ y_{11}^{11}-f}{P_2} =11(x_{11}y_{11}-1) \tag{4}$$ where $P_1,P_2$ are polynomials in $x_{11},y_{11}$ analogous to the previous ones?
P.S. Just a remark. These relationships can be proven after knowing about it, but how in the world does Ramanujan come up with them in the first place?
After staring at my question for a while, I realized how to answer it. Let,
$$a,b,c=z_1^{1/11}, z_2^{1/11}, z_3^{1/11}$$
hence (for convenience), $p=a+b+c=x_{11},\; q=ab+ac+bc=y_{11},\; abc = 1$. Then, $$\frac{x_{11}^{11}-e}{P_1} = 11(x_{11}y_{11}-1)$$ is equivalent to, $$(a+b+c)^{11}-(a^{11}+ b^{11}+ c^{11}) = 11(pq-1)\, P_1$$ Since $pq-1 = (a+b)(a+c)(b+c)$ if $abc=1$, then all that remains is to express $P_1$ in terms of $p,q$. One finds that, $$P_1 = (p^2 - q) (p^6 - 3 p^4 q + 4 p^2 q^2 - q^3) + p (3 p^2 - 5 q) (p^2 - q) + (2p^2 - q)$$ holds true if $abc=1$. For $P_2$, let $p,q\to q,p$.
I used Mathematica to find this, and the more complicated form is probably the reason why Ramanujan didn't include it in his Notebooks. Even though he didn't write "...and so on" indicating there were more identities, he probably did know.
P.S. I didn't check if it can be done for $n=9$, though as T. Andrews points out, it may work for odd n.