Asymptotic expansion in a polinomial The problem is about Perturbation Theory

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*Find the first-order approximation for the roots of:
$$P^\epsilon(x)=\epsilon^2x^6-\epsilon x^4-x^3+8=0 $$ Here
you no longer know the exact solution by radicands. How can you
validate your asymptotic approximation? Compare where you can by
setting a value of $\epsilon$.

I tried with dominant balance with the diferents variable but the only way is viable propose is
$x=y/\delta(\epsilon)$ with $\delta(\epsilon)=\epsilon$, but i find only one root and graphically there are two reals roots for $\epsilon<<1$.
Someone can help me? :(
 A: Regular perturbation case: $x \sim c_1$ i.e. bounded but not small, and you get a perturbation of the unperturbed root $2$.
Singular perturbation case: choose $x \sim c_1 \epsilon^a$ with $a<0$ so that the two terms with largest negative exponent cancel out. Thus the question is what is biggest for negative $a$: $2+6a,1+4a,3a$? It depends what $a$ is, but you can sweep through all the cases where a cancellation is possible:
$$2+6a=1+4a \Rightarrow a=-1/2 \text{ retaining } -1, \text{ neglecting } -3/2 \\
2+6a=3a \Rightarrow a=-2/3 \text{ retaining } -2, \text{ neglecting } -5/3 \\
1+4a=3a \Rightarrow a=-1 \text{ retaining } -3,\text { neglecting } -4.$$
Only the second one is consistent, i.e. the terms that you are cancelling are actually the dominant terms. So this determines the correct $a$. And indeed if you use software to find the root for $\epsilon=10^{-10}$ and $\epsilon=10^{-40}$, you find they differ by about a factor of $10^{20}=10^{2/3 (30)}$.
This surprised me a little bit because it indicates that $\epsilon x^4$ only changes the large root by a regular perturbation!
