You are not supposed to find two sets whose union is $\mathbb{R} $ so your statement about making a cut on the real line does not seem relevant here. You should first write down the definition of Dedekind cut and verify each of the properties for the given set $A=\{r\in\mathbb{Q},r^3<2\} $. Most books list the following properties for Dedekind cut $A$:
- $\emptyset \neq A\neq\mathbb {Q} $ and $A\subset\mathbb {Q} $.
- If $r, s$ are rational numbers such that $r>s$ and $r\in A$ then $s\in A$ ie if some rational number lies in $A$ then all smaller rationals also lie in $A$.
- $A$ does not possess a maximum element ie if $r\in A$ then there is an element $s\in A$ with $r<s$.
Now you need to verify the above properties for your given set $A$. The first property is obvious because $1\in A, 2\notin A$. For second property let $r\in A$ and $r>s$ then we can see that $s^3<r^3<2$ so that $s\in A$.
It is the third property which requires some effort to verify. You have to show that if $r\in A$ then there is another $s\in A$ with $r<s$. Since $r\in A$ we have $r^3<2$ and then $d=2-r^3>0$. We will use $r, d$ to find a number $s\in A$ with $r<s$. Since we want $r<s$ we can first try writing $s=r+h$ where $h>0$ and we need to find $h$ suitably so that $s=r+h\in A$ ie $(r+h) ^3<2$ ie $$r^3+3r^2h+3rh^2+h^3<2$$ or $$h(3r^2+3rh+h^2)<2-r^3=d$$ Next let's choose $h<1$ so that $$3r^2h+3rh+h^2<3r^2+3r+1$$ ie $$h(3r^2+3rh+h^2)<h(3r^2+3r+1)$$ If in addition to $h<1$ we also ensure that $h<d/(3r^2+3r+1)$ then we have the desired inequality $(r+h) ^3<2$. Thus we just need to take $h$ less than $\min(1,d/(3r^2+3r+1))$ and then $s=r+h\in A$ and thus $A$ has no maximum element.