= Solution
Take $\epsilon\to0^+$. A root on the smaller scale has $x=\epsilon u$, giving $u+2+\epsilon u^3=0$. Put $u=u_0+\epsilon u_1+O(\epsilon^2)$. Successive <coefficients> give $u_0=-2$ and $u_1+u_0^3=0$, hence $u_1=8$. The real root is therefore
$$
\boxed{x_0=-2\epsilon+8\epsilon^2+O(\epsilon^3).}
$$
The other <dominant balance> is between the cubic and linear terms, requiring $x=\epsilon^{1/2}z$. Set $t=\epsilon^{1/2}$; the equation becomes $z^3+z+2t=0$. At $t=0$ the two nonzero roots are $z_0=\pm i$. With $z=z_0+t z_1+O(t^2)$, the first correction satisfies $(3z_0^2+1)z_1+2=0$, so $-2z_1+2=0$ and $z_1=1$. Thus
$$
\boxed{x_+=i\epsilon^{1/2}+\epsilon+O(\epsilon^{3/2}),\qquad
x_-=-i\epsilon^{1/2}+\epsilon+O(\epsilon^{3/2}).}
$$
The limiting roots in both rescalings are simple, so the <implicit function theorem> justifies these regular expansions in their respective small parameters. The fractional powers in the original variable are a <fractional-power root expansion>. The real <derivative> $3x^2+\epsilon$ is strictly positive; the cubic therefore has exactly one real root. Its other roots are the displayed <complex conjugate> pair, accounting for all three roots. Their leading real corrections sum to $-2\epsilon+\epsilon+\epsilon=0$, as required by the absence of a quadratic <coefficient>.
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