Automorphisms of transcendental extensions. I'm trying to solve the following two exercises from chapter 18 of Garling's A Course in Galois Theory.



I haven't been able to get started (only thing I've done is verify that $\sigma^3$ is the identity, and calculated some examples that me guess the fixed field of $\sigma$ is K). The chapter of the book where this is taken from gives few hints on where to start to tackle these as it only mentions some theorems about algebraic independence (which won't be aplicable here as the extension is simple) and the fact that $K(\alpha)$ is $K(x)$ if $\alpha$ is transcendetal over $K$ which I already knew and its the way I think about $K(\alpha)$ but I haven't been able to put it to good use. 

Any help would be greatly appreciated.
I use the variable $t$ instead of $\alpha$ since its easier type.

Clearly the elementary  symmetric functions in $t,\sigma(t), \sigma^2(t)$ belong to the fixed field. And further $t$ is the root of a cubic polynomial in these functions. If you calculate them you get only one quantity,
$$s=\frac{t^3-3t+1}{t(1-t)}$$ 

Thus $k(t)/k(s)$ is an extension of degree $3$ with $\sigma$ as automorphism.
Thus the fixed field is $k(s)$.
I use the variable $t$ instead of $\alpha$ since its easier type.

Clearly the elementary  symmetric functions in $t,\sigma(t), \sigma^2(t)$ belong to the fixed field. And further $t$ is the root of a cubic polynomial in these functions. If you calculate them you get only one quantity,
$$s=\frac{t^3-3t+1}{t(1-t)}$$ 

Thus $k(t)/k(s)$ is an extension of degree $3$ with $\sigma$ as automorphism.
Thus the fixed field is $k(s)$.