= Solution
Use <dominant balance for algebraic roots> to separate the three scales. On writing $x=\epsilon X$, the equation becomes $X^3-X^2+2\epsilon X+2\epsilon^3=0$. Its nonzero simple limiting root is $X=1$, so one root has $x_1\sim\epsilon$.
For the two remaining roots put $x=\epsilon^2Y$. Division by $\epsilon^5$ gives $\epsilon Y^3-Y^2+2Y+2\epsilon=0$. The nonzero limiting root is $Y=2$, giving $x_2\sim2\epsilon^2$. Finally $x=\epsilon^3Z$ gives, after division by $\epsilon^6$,
$$
\epsilon^3Z^3-\epsilon Z^2+2Z+2=0.
$$
Its simple limiting root is $Z=-1$. These three distinct balances account for all roots of the cubic, and
$$
\boxed{x_1\sim\epsilon,\qquad x_2\sim2\epsilon^2,\qquad x_3\sim-\epsilon^3.}
$$
For the root of smallest magnitude, set $Z=-1+b\epsilon+O(\epsilon^2)$. The order-$\epsilon$ coefficient is $-1+2b$, so $b=1/2$. Hence
$$
\boxed{x_3=-\epsilon^3+\tfrac12\epsilon^4+O(\epsilon^5).}
$$
Each rescaled root is simple at $\epsilon=0$, justifying the <regular perturbation of a simple algebraic root> rather than treating the original triple root as an ordinary regular perturbation.
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