I have a cubic equations with 2 unknowns and constraints on both variables. I am unsure whether or not a closed form solution does exist. How should I go about solving this?
${ -1 + 2b - 2b^2 - \frac{5}{6}b^3 + 2e - e^2 + \frac{e^3}{6} - 2eb + \frac{e^2b}{2} = 0 }$
or alternatively (not developed)
${ \frac{1}{3} - b^2(1 + b + \frac{e}{2}) + \frac{1}{6}(b^3-(-2 -b -e)^3) = 0 }$
given that ${ e \in [0,1]}$ and ${ b \in [-1,0] }$
Based on the comments, it appears you are trying to maximize $e$, subject to the given constraints.
Trying $e=1$, the cubic equation reduces to $$5b^3+12b^2-3b-1=0$$ which has $3$ real solutions for $b$, one of which, $b\approx -.1944051955$, satisfies the given constraints.
Therefore the maximum value of $e$, subject to the given constraints, is $e=1$.