If I am originally working with $\mathbb{F} = \mathbb{Z}/n \mathbb{Z}$, with $n$ prime, and if I add both the values $\sqrt[3]{x}$ and $\left( \sqrt[3]{x} \right)^2$, will this create a field with $n^3$ elements?
What if I add $\sqrt[2]{x}$ and $\sqrt[3]{x}$ to the field $\mathbb{Z}/n \mathbb{Z}$? How many elements can I get out of this field, assuming that it's a field?
I'm essentially trying to find a field that has an element of multiplicative order $n^3-1$ by adjoining elements to the field $\mathbb{Z}/n \mathbb{Z}$. The real problem is that I don't want to use $\mathbb{F[y]}/p(y)$ for some polynomial in $y$ - in other words, I don't want to factor any polynomials to find this field.
Instead of thinking about which "elements" you are adjoining to the field, think about what properties this new element has. For example when take $\mathbb{C}=\mathbb{R}[i]$ and construct the complex numbers we are adjoining a solution to the equation $x^2+1=0$. If there is no polynomial equation that puts the new element back into the old field then we get an infinite extension. For example $\mathbb{R}[x]$ gives us rational functions.
To get a finite new ring from a finite old ring we need to adjoin something that satisfies a polynomial relation putting it back into the old ring. So for example given the field $\mathbb{Z}/2\mathbb{Z}$ we can easily test that the equation $x^2+x+1=0$ has no solutions ($1+1+1=1$, and $0+0+1=1$). Thus we can get a new field buy taking $\mathbb{Z}/2\mathbb{Z}[\alpha]$ where we simply dictate that the new element satisfies $\alpha^2+\alpha+1=0$.
It turns out there is a very simple way of determining the size of the new field given the size of the old field. If $d$ is the degree of the polynomial relation that $\alpha$ satisfies (where we take the smallest polynomial we can that gets back into the original field) then $$|\mathbb{Z}/p\mathbb{Z}[\alpha]|=|\mathbb{Z}/p\mathbb{Z}|^d$$ So to cube the size of the field, we need to find a cubic equation that no existing elements satisfy, and adjoin a solution.