If $\alpha, \beta$ and $\gamma$ are the roots of $x^3-9 x^2+26 x-24=0$ such that $\alpha=2 \beta$.
Clearly, $\alpha+\beta+\gamma=9, \alpha \beta+\beta \gamma+\gamma \alpha=-26$ and $\alpha \beta \gamma=24$
$$
\begin{array}{lr}
\therefore & 3 \beta+\gamma=9 \text { and } \beta^2 \gamma=12 \\
\Rightarrow & \beta^2(9-3 \beta)=12 \Rightarrow 9 \beta^2-3 \beta^3=12 \\
\Rightarrow & \beta^3-3 \beta^2+4=0 \\
\Rightarrow & \left.\beta+1) \beta^2-4 \beta+4\right)=0 \\
\Rightarrow & \beta+1) \beta-4^2=0 \Rightarrow \beta=-1 \text { or } \beta=2
\end{array}
$$
When $\beta=-1$, then $\alpha=-2$ and in this case
$$
\alpha^3+\beta^3=-8-1=-9
$$
When $\beta=2$ then $\alpha=4$ and in this case
$$
\alpha^3+\beta^3=64+8=72
$$