Let \(x=\sqrt[3]{a+b\sqrt{c}}\) and \(y=\sqrt[3]{a-b\sqrt{c}}\), and \(x+y=1\). Cube both sides: \[ \begin{align*} x^3+y^3+3xy(x+y)&=1 \\ 2a+3xy&=1 \\ xy&=\frac{1-2a}{3} \end{align*}\] This means \(xy \in \mathbb{Q}\), and \((xy)^3=a^2-b^2c \in \mathbb{Z}\), therefore \(xy\in \mathbb{Z}\).
Let \(xy=-n\). Then \(2a-1=3n\), so \(a=(3n+1)/2\). Since \(a\) is integer, \(n\) is odd.
Now set \(n=2m-1\). Then \(a=3m-1\). So, \[ \begin{align*} (-n)^3&=(3m-1)^2-b^2c \\ b^2c&=(3m-1)^2+(2m-1)^2 \\ b^2c&=m^2(8m-3) \end{align*}\] Now write, \(8m-3=r^2s\), where \(s\) is squarefree. Then \[b^2c=(mr)^2s.\] So every solution comes from choosing a divisor \(d \mid mr\), then setting \(b=mr/d\) and \(c=sd^2\).
Therefore count all paris \((m, d)\) such that \(d \mid mr\) and: \[3m-1+mr/d+sd^2\leq 110 000 000.\]