Method of dominant balance and perturbation Approximate the solutions of $$\epsilon x^4 + (x-1)^3=0$$

I can't perform a singular perturbation because if I let $\epsilon=0$ then I lose a root.  My professor suggests The Method of Dominant Balance for problems like this, but I don't really know how to proceed with that method.

If someone could just get me started, then I could do the rest in maple.  

Thanks
If we naively just send $\epsilon \to 0$ then we get the equation

$$
(x-1)^3 = 0,
$$

so we can deduce that we have three roots tending to $x=1$ as $\epsilon \to 0$.  We'll suppose they have asymptotic series of the form

$$
x \approx 1 + \sum_{k=1}^{\infty} a_k \delta_k(\epsilon),
$$

where

$$
1 \gg \delta_1(\epsilon) \gg \delta_2(\epsilon) \gg \delta_3(\epsilon) \gg \cdots
$$

as $\epsilon \to 0$.

If we substitute just the first two terms $x \approx 1 + a_1 \delta_1(\epsilon)$ into the original equation

$$
\epsilon x^4 + (x-1)^3 = 0 \tag{$*$}
$$

and expand it we get

$$
a_1^4\epsilon\delta_1(\epsilon)^4 + 4a_1^3\epsilon\delta_1(\epsilon)^3 + a_1^3\delta_1(\epsilon)^3 + 6a_1^2\epsilon\delta_1(\epsilon)^2 + 4a_1\epsilon\delta_1(\epsilon) + \epsilon \approx 0.
$$

Now we'll apply the method of dominant balance.  First, certain terms are, by assumption, smaller than others, so they could not possibly be part of a dominant balance and may be ignored.  Specifically we know that

$$
a_1^4\epsilon\delta_1(\epsilon)^4 \ll 4a_1^3\epsilon\delta_1(\epsilon)^3 \ll 6a_1^2\epsilon\delta_1(\epsilon)^2 \ll 4a_1\epsilon\delta_1(\epsilon) \ll \epsilon,
$$

so if we ignore all but the largest of these then our equation becomes

$$
a_1^3\delta_1(\epsilon)^3 + \epsilon \approx 0.
$$

From this we see that

$$
\delta_1(\epsilon) = \epsilon^{1/3} \quad \text{and} \quad a_1^3 = -1,
$$

and with three choices for $a_1$---namely the three cube roots of $-1$---we obtain approximations for each of the three roots of the original equation with tend to $x=1$:


  
  $x \approx 1 - \epsilon^{1/3}$,
  $x \approx 1 + e^{i\pi/3} \epsilon^{1/3}$, and
  $x \approx 1 + e^{i5\pi/3} \epsilon^{1/3}$.
  


This suggests that we might be able to take

$$
\delta_k(\epsilon) = \epsilon^{k/3}, \quad k \geq 1,
$$

and indeed if we substitute

$$
x \approx 1 + \sum_{k=1}^{\infty} a_k \epsilon^{k/3}
$$

into $(*)$ and collect like powers of $\epsilon$ then we obtain equations for the coefficients $a_k$,


$a_1^3 + 1 = 0$,
$3 a_2 a_1^2+4 a_1 = 0$,
$3 a_3 a_1^2+6 a_1^2+3 a_2^2 a_1+4 a_2 = 0$, etc.


It just remains to find the last root of $(*)$.  If we simply expand it out we get

$$
\epsilon x^4 + x^3 - 3x^2 + 3x - 1 = 0. \tag{$**$}
$$

We will again use the method of dominant balance.  By ignoring the $\epsilon x^4$ term in the very beginning we essentially assumed it would not be part of a dominant balance, so to find the last root we must assume the opposite.  As such we'll be looking for balances between $\epsilon x^4$ and the remaining four terms, so we'll consider the cases


$\epsilon x^4 \asymp 1$,
$\epsilon x^4 \asymp 3x$,
$\epsilon x^4 \asymp 3x^2$, and
$\epsilon x^4 \asymp x^3$.


In case 1 we have $x \asymp \epsilon^{-1/4}$, but then the term $3x$, being $\asymp \epsilon^{-1/4}$, dominates the terms $\epsilon x^4$ and $1$ in the balance, so the balance is not dominant.

In case $2$ we have $x \asymp \epsilon^{-1/3}$, but then the term $-3x^2$, being $\asymp \epsilon^{-2/3}$, dominates both terms in the balance, so the balance is again not dominant.

In case $3$ we have $x \asymp \epsilon^{-1/2}$, but then the term $x^3$, being $\asymp \epsilon^{-3/2}$, dominates both terms in the balance, namely $\epsilon x^4$ and $x^2$, which are $\asymp \epsilon^{-1}$.

In case $4$ we have $x \asymp \epsilon^{-1}$, and this balance is dominant since

$$
\epsilon x^4 \asymp x^3 \asymp \epsilon^{-3} \gg 3x^2 \gg 3x \gg 1.
$$

Ignoring the terms outside of the balance we then have

$$
\epsilon x^4 + x^3 \approx 0,
$$

so that


  $$
x \approx - \epsilon^{-1}.
$$


We might then suspect that this root has an asymptotic series of the form

$$
x \approx -\epsilon^{-1} + \sum_{k=0}^{\infty} a_k \epsilon^k,
$$

and if we substitute this into $(*)$ or $(**)$ and collect like powers of $\epsilon$ we obtain equations for the coefficients $a_k$,


$a_0+3 = 0$,
$3 a_0^2+6 a_0-a_1-3 = 0$,
$-3 a_0^3-3 a_0^2+6 a_1 a_0+3 a_0+6 a_1-a_2-1 = 0$, etc.


We've thus accounted for all four roots of the polynomial equation and can calculate as many terms of their asymptotic series as we wish.
Let $\epsilon=\tau^3$, then we can solve the equation

$$0=\tau^3 x^4 + (x-1)^3$$

There are 4 roots.  When $\tau \to 0$, they are given by:

$$x_1=-3-\frac{1}{\tau^{3}}$$
$$x_2=1-\tau$$
$$x_3=1+\left(\frac{1}{2}-i\frac{1}{\sqrt{3}}\right)\tau$$
$$x_4=1+\left(\frac{1}{2}+i\frac{1}{\sqrt{3}}\right)\tau$$