= Solution
Write the equation as $\epsilon x^3-(x-1)^2=0$. For $\epsilon\to0^+$, two <roots of a polynomial> approach the double root one. Their splitting follows a fractional-power <dominant balance>, rather than an ordinary expansion in integer powers.
Put $\delta=\sqrt\epsilon$ and $x=1+a\delta+b\delta^2+\cdots$. Substitution gives
$$
0=(1-a^2)\delta^2+(3a-2ab)\delta^3+O(\delta^4).
$$
Thus $a=\pm1$ and $b=3/2$. The two bounded roots have the <Puiseux series> expansions
$$
\boxed{x_\pm=1\pm\sqrt\epsilon+\frac32\epsilon+O(\epsilon^{3/2}).}
$$
The first two nonzero terms are $1\pm\sqrt\epsilon$; the displayed correction also resolves their common order-$\epsilon$ shift.
The third root escapes to infinity because the leading <polynomial degree> drops when $\epsilon=0$. Balance $\epsilon x^3$ against $x^2$, giving $x=O(\epsilon^{-1})$. Set $X=\epsilon x$; the equation becomes
$$
X^3-X^2+2\epsilon X-\epsilon^2=0.
$$
For $X=1+A\epsilon+B\epsilon^2+\cdots$, the first two coefficient equations are $A+2=0$ and $B+2A^2+2A-1=0$. Hence
$$
\boxed{x_\infty=\epsilon^{-1}-2-3\epsilon+O(\epsilon^2).}
$$
Its first two nonzero terms are $\epsilon^{-1}-2$. These three distinct branches account for all roots of the cubic for sufficiently small positive $\epsilon$. The <Vieta formulas> provide a consistency check: their sum is $\epsilon^{-1}$, with the constant and order-$\epsilon$ terms cancelling.
If $\epsilon$ approaches zero through negative values, the bounded pair is complex: $x_\pm=1\pm i\sqrt{|\epsilon|}+\tfrac32\epsilon+O(|\epsilon|^{3/2})$. The large-root expansion is unchanged as a real asymptotic formula. The positive-parameter regime is used for the boundary-layer problem below.
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