Let α, β, γ are positive real numbers such that _γ(2 α)=(1)/(3),...

$\gamma=8 \alpha^3, \quad \gamma=5^6 \beta^6, \quad \gamma^3=\alpha^2 \beta^2$$\begin{array}{ll}\therefore & (2 \alpha)^3=\left(25 \beta^2\right)^3 \\ \therefore & 2 \alpha=25 \beta^2\end{array}$$Also$\quad $\gamma^3=2^9$\cdot \alpha^9=$\alpha^2$\cdot \beta^2 \quad \Rightarrow \quad $\beta^2=2^9$\cdot \alpha^7=$\frac{2 \alpha}{25}\therefore \quad \alpha^6=$\frac{1}{2^8 \cdot 25}=\frac{1}{2^8 \cdot 5^2} \quad \Rightarrow \quad \alpha^3=$\frac{1}{2^4 \cdot 5} \quad Option (2) Also$\quad $\gamma=$8 $\cdot \frac{1}{2^4 \cdot 5}=\frac{1}{10} \quad Option (3) Also$\quad $\beta^6=$\frac{\gamma}{5^6}=\frac{1}{10 \cdot 5^6}