two-term asymptotic expansion for each root:
$\epsilon z^8-z^3-1=0$ as $\epsilon \rightarrow 0$
how to find other solutions than the one near -1?
first, I check when $\epsilon=0$, I get the root of $z=-1$, though I am not sure if this root is of order 1 or order 3 (because $z^3=-1$). In this way, I can easily obtain a root that is near $z=-1$.
Next, I try to balance $\epsilon z^8 \sim z^3$ and write that $z=\epsilon^{-1/5}+\tilde{z}$ with $\tilde{z} \ll \epsilon^{-1/5}$ and plug it back in the polynomial equation. By doing this, I obtain a new 3rd order polynomial equation for $\tilde{z}$. And I am not sure what to do next. Is it the correct way to solve this kind of problem?
Thanks
I shall assume $\epsilon >0$ to make (my) life easier.
Consider that, in the real domain, you look for the zero of function $$f(z)=\epsilon z^8-z^3-1$$ As, you wrote, if $\epsilon$ is small, one of the roots is close to $z=-1$.
Expand $f(z)$ as a Taylor series around $z=-1$ to get
$$f(z)=\epsilon - (8 \epsilon +3)(z+1)+ (28 \epsilon +3)(z+1)^2+O\left((z+1)^3\right)$$ and make a series reversion which would give $$z=-1-\frac{f(z)-\epsilon }{8 \epsilon +3}+\frac{(28 \epsilon +3) (f(z)-\epsilon )^2}{(8 \epsilon +3)^3 }+O\left((y-\epsilon )^3\right)$$ Since what is desired is $f(z)=0$, then an approximation is $$z=-1+\frac{\epsilon }{8 \epsilon +3}+\frac{(28 \epsilon +3) \epsilon ^2}{(8 \epsilon +3)^3 }\tag 1$$ Now, you could continue using Taylor series around $\epsilon=0$ and have the simpler $$z=-1+\frac{\epsilon }{3}-\frac{7 \epsilon ^2}{9}+\frac{68 \epsilon ^3}{27}+O\left(\epsilon ^4\right)\tag 2$$ what you could have obtained by a series solution with binomial expansions.
Now, what about the other roots ?
We have $$f'(z)=8 \epsilon z^7 -3 z^2 \qquad f''(z)=56 \epsilon z^6 -6 z$$ In the real domain, the first derivative cancels at $$z_*=\left(\frac{3}{8 \epsilon}\right)^{\frac{1}{5}}$$ for which $$f(z_*)=-1-\frac 5{16}\left(\frac{27}{16 \epsilon^2}\right)^{\frac{1}{5}}\quad < ~~ 0 $$ $$f''(z_*)=15\left(\frac{3}{8 \epsilon}\right)^{\frac{1}{5}}\quad > ~~ 0$$ So, there is only one real root which is larger than $z_*$.
So, expand $f(z)$ around $z_*$ $$f(z)=f(z_*)+\frac 12 f''(z_*) (z-z_*)^2+O((z-z_*)^3$$ Ignoring the higher order terms $$z=z_*+\sqrt{-2\frac {f(z_*)}{f''(z_*)}}$$ Expanding again as a series $$z=\frac{6+\sqrt{3}}{2\sqrt[5]{648}} {\epsilon^{-\frac 15}}+O\left(\epsilon^{\frac 25} \right)\tag 3$$
Now, let us chek for $\epsilon=10^{-k}$
$$\left( \begin{array}{ccc} k & \text{estimate } (2) & \text{"exact" solution} \\ 1 & -0.971925925925926 & -0.972570683163445 \\ 2 & -0.996741925925926 & -0.996741960146079 \\ 3 & -0.999667441925926 & -0.999667441874155 \\ 4 & -0.999966674441926 & -0.999966674441865 \\ 5 & -0.999996666744442 & -0.999996666744442 \end{array} \right)$$
For the largest root, it is less spectacular but it seems t $$\left( \begin{array}{ccc} k & \text{estimate } (3) & \text{"exact" solution} \\ 1 & 1.6786096 & 1.6499102 \\ 2 & 2.6604169 & 2.5417606 \\ 3 & 4.2164767 & 3.9934957 \\ 4 & 6.6826652 & 6.3145773 \\ 5 & 10.591311 & 10.001998 \end{array} \right)$$