ca.analysis and odes - Asymptotic series for roots of polynomials

If you have a good initial guess (such as your case), then you don't need either undetermined coefficients or Newton's method (although they both work), you can in fact get a procedure which will give you as many coefficients as you want. Harald's first step is excellent, so we'll start from that. Given $F(epsilon,w) = epsilon^2w+epsilon w^2+w^3-1 = 0$, we can derive a differential equation for $w(epsilon)$. With the help of Maple's gfun[algeqtodiffeq], giving
$$ 9+ left( 6{epsilon}^{5}+7{epsilon}^{2} right) w ( epsilon ) + left( 27-6{epsilon}^{6}-7{epsilon}^{
3} right) {frac {d}{depsilon}}w(epsilon) + left( -3{epsilon}^{7}-14{epsilon}^{4}-27epsilon right) {frac {d^{2}}{d{epsilon}^{2}}}w ( epsilon ) $$
with initial conditions $w (0) =1,w^{'''}(0) =-4/9 $ [the DE is singular at $epsilon=0$ so the initial conditions are non-standard). Nevertheless, from there one can continue and use gfun[diffeqtorec] to get a recurrence for the coefficients of the series. The result is
$$ left( 6-3n-3{n}^{2} right) u(n) + left( -98-77n-14{n}^{2} right) u (n+3) + left( -
648-27{n}^{2}-270n right) u (n+6 ) $$
with starting conditions $u(0) =1, u(1) =-1/3,u (2) =-2/9,u (3) ={frac {7}{81}},u(4) =0,u(5) ={frac {14}{729}}$. These are sufficient to allow you to 'unwind' the recurrence as much as you want.

Bjorn's answer gives the underlying 'analytic' answer that justifies that the above gives a convergent result. Actually, the procedure above works in other cases as well, but then the result is only valid in a sector, and computing the size of that sector can be fiendishly difficult.