 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 $d_i<c_i$. Sinceminimality 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 $3{\lambda }_i^2\ne 0$. 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