= Solution
The first characteristic <polynomial> factors as
$$
\rho(\zeta)=\zeta^3+\alpha\zeta^2-\alpha\zeta-1
=(\zeta-1)[\zeta^2+(\alpha+1)\zeta+1].
$$
The two quadratic roots have product one. Both can lie in the closed unit disk only if both have modulus one, requiring $|\alpha+1|\leq2$. At $\alpha=-3$ the root one is triple and $\beta=0$; at $\alpha=1$ the root minus one is double. Both fail the <root condition for a multistep method>. In the intervening open interval the two quadratic roots are distinct conjugate unit-modulus roots, different from one. Thus
$$
\boxed{-3<\alpha<1\quad\text{is exactly the zero-stability range}.}
$$
Using $2\beta=\alpha+3$, the supplied expansion gives
$$
C_0=C_1=C_2=0,\qquad C_3=\frac{9-\alpha}{12},\qquad
C_4=\frac{9-\alpha}{8},\qquad C_5=\frac{231-23\alpha}{240}.
$$
Throughout the <zero-stable> range, $C_3\ne0$, so \b[every such method has order two]. There is a unique higher-order choice:
$$
\boxed{\alpha=9,\qquad\beta=6.}
$$
Here $C_3=C_4=0$ and $C_5=1/10\ne0$, giving exactly order four, with normalized leading truncation error $h^4y^{(5)}/120$. But the extra roots are $-5\pm2\sqrt6$, one of modulus greater than one. By the <Dahlquist equivalence theorem> this consistent but non-<zero-stable> method is \b[not convergent] for arbitrary consistent starting values.
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