I would suggest using some Newton-Raphson scheme.
1. Let f(x)=x^{3}-a
You are to find X so that f(X)=0.
2. Pick an initial value x_{0}\to{f}(x_{0})=x_{0}^{3}-a
3. The equation for the tangent line L(x)=at (x_{0},f(x_{0}) is given by:
L(x)=f(x_{0})+f'(x_{0})(x-x_{0})
4- Let the next iteration point be the x-intercept of L(x):
L(x_{1})=0\to{x}_{1}=x_{0}-\frac{f(x_{0})}{f'(x_{0}}
5. Or, in this case, the iterative scheme becomes:
x_{n}=x_{n-1}-\frac{x_{n-1}-\frac{a}{x_{n-1}^{2}}}{3}
That is:
x_{n}=\frac{2x_{n-1}^{3}+a}{3x_{n-1}^{2}}, n\geq{1}