We can use Taylor series to approximate $a + \frac{b}{3a}\approx \sqrt[3]{a^3 + b}$ with $b \ll a$.
However, what are precise upper and lower bounds for this quantity?
$$ a + \frac{b}{3a^2} < \sqrt[3]{a^3 + b} < a + \frac{b}{3a^2} + \dots$$
I don't know numerical methods, so I don't know how to say how "good" or "close" these approximations are...
I can imagine the accuracy goes are $error \propto b^k $ for some $k$.
EDIT I repeat I don't know how to make this more precise, so I give two ideas. Let $|\epsilon| < M$ then what is the smallest $k$ such that
$$1 +\tfrac{1}{3}\epsilon < \sqrt[3]{1 + \epsilon} < 1 + k(M)\epsilon$$
for all "allowable" $\epsilon$ i.e. $|\epsilon| < M$.
Another idea is to look at my previous response where I find the rather peculiar inequalities:
$$ a + \frac{ab}{3a^3 + b} < (a^3 + b)^{1/3} < \frac{a}{2} + \sqrt{\frac{a^2}{4} + \frac{b}{3a}} $$
Estimates like these are less practical, and come from a time when calculators were not available.