= Solution
If the nonzero $\lambda_i$ are real and distinct, their cubes are also distinct. Suppose the multiplicity of $x-\lambda_i^3$ in $m_{\alpha^3}$ were $d_i<c_i$. Since
$$
m_{\alpha^3}(\alpha^3)=0,
$$
minimality makes $m_\alpha(x)$ divide $m_{\alpha^3}(x^3)$. At $x=\lambda_i$, the factor $x^3-\lambda_i^3$ has a simple zero because its derivative is $3\lambda_i^2\ne0$. Thus $m_{\alpha^3}(x^3)$ has multiplicity only $d_i$ there, contradicting the multiplicity $c_i$ in $m_\alpha$. Every exponent is therefore $c_i$, and
$$
\boxed{m_{\alpha^3}(x)=
\prod_{i=1}^k(x-\lambda_i^3)^{c_i}}.
$$
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