If we are given that $F(x) = x-x^3$ is a force function, which means it is in the units of $[M][L][T]^{-2}$, then how do we determine what kind of "unit units" participate in this function?
Namely, we have: $[x-x^3]=[M][L][T]^{-2} \implies [x][1-x^2]=[M][L][T]^{-2}$, so we have unit mass $m$: $[m][x][1-x^2]=[M][L][T]^{-2}$. We also observe that the $[x]$ corresponds to $[L]$. Now, there should be $[T]^{-2}$ somewhere in LHS, so we can deduce something like this for a unit function of time $t$: $[m][x][1/t^2][1-x^2]=[M][L][T]^{-2}$ or $[m][a][1-x^2]=[M][L][T]^{-2}$, where $a$ is unit acceleration. We should also probably deduce that there is a unit of $[L]^{-2}$ before the $x^2$ in $[1-x^2]$, so we thus have $[m][a]([1]-[1/d^2][x^2])=[M][L][T]^{-2}$, where $1/d^2$ is some unit function of distance.
I think that I'm not quite correct in some of my reasoning, so I'd appreciate some insight.
In an algebraic expression, all terms which are added or subtracted must have the same dimensions. This implies that each term on the left-hand side of an equation must have the same dimensions as each term on the right-hand side.
So you have that $$ [F]=[x]=[x^3]=[M][L][T]^{-2} $$ It's obvious thatt if $x$ and $x^3$ have the same dimension, there exist at least two unitary constant $\alpha=1$ and $\beta=1$ such that $[\alpha x]=[\beta x^3]=[M][L][T]^{-2}$, that is \begin{align} [\alpha] [x]&=[M][L][T]^{-2}\\ [\beta] [x]^3&=[M][L][T]^{-2} \end{align} If you choose $x$ as lenght, that is $[x]=[L]$, you'll have $[\alpha]=[M][T]^{-2}$ and $[\beta]=[M][L]^{-2}[T]^{-2}$. If you choose $x$ as acceleration, that is $[x]=[L][T]^{-2}$, you'll have $[\alpha]=[M]$ and $[\beta]=[M][L]^{-2}[T]^{4}$. and so on.