Sextic twist optimization of BN pairing - cubic root extraction required?
You need to recall how the extension is built. $\mathbb{F}_{p^{12}}$ is built on top of $\mathbb{F}_{p^2}$ using the reduction polynomial $f(x) = x^6 - \xi$, where $\xi \in \mathbb{F}_{p^2}$ is a non-square and non-cube (using the notation from the paper). In other words, this is the set of polynomials with coefficients in $\mathbb{F}_{p^2}$, modulo $f(x)$.

Since we're working modulo $f(x)$, then $f(x) = 0$ which means that you can write $x = \sqrt[6]{\xi}$. However, this does not mean that you actually carry out the sextic root (it does not exist in $\mathbb{F}_{p^2}$), but you can "define" the sextic root of $\xi$ as $x$ (which is an element of $\mathbb{F}_{p^{12}}$: $x = 0x^5 + 0x^4 + 0x^3 + 0x^2 + 1x^1 + 0$). This process is called "adjoining roots": here is a good explanation. (A good analogue is with how you build the complex numbers by "defining" $i$ as the square root of $-1$, which does not exist in the reals).

In your question, $i = \xi$. This means that $i^{1/2} = (i^{1/6})^3 = x^3$ and $i^{1/3} = (i^{1/6})^2 = x^2$. They are just $\mathbb{F}_{p^{12}}$ elements with one coefficient $1$ and the others are zero and are trivially "computed"; you don't need to carry out any square or cube root.

There are some other issues since the sextic extension is often built as a quadratic extension over a cubic extension, but the reasoning is the same.