If the co-efficient of $$x^9$$ in $$\left(\alpha x^3 + \frac{1}{\beta x}\right)^{11}$$ and the co-efficient of $$x^{-9}$$ in $$\left(\alpha x - \frac{1}{\beta x^3}\right)^{11}$$ are equal, then $$(\alpha\beta)^2$$ is equal to

Since the coefficient of $$x^9$$ in $$\left(\alpha x^3 + \frac{1}{\beta x}\right)^{11}$$ equals the coefficient of $$x^{-9}$$ in $$\left(\alpha x - \frac{1}{\beta x^3}\right)^{11}$$, we start by finding the former. Considering the general term in the expansion of $$\left(\alpha x^3 + \frac{1}{\beta x}\right)^{11}$$:
$$T_{r+1} = \binom{11}{r}(\alpha x^3)^{11-r}\left(\frac{1}{\beta x}\right)^r = \binom{11}{r}\frac{\alpha^{11-r}}{\beta^r} x^{33-4r}.$$
Substituting $$33 - 4r = 9$$ gives $$r = 6$$, so the coefficient of $$x^9$$ is $$\binom{11}{6}\frac{\alpha^5}{\beta^6}$$.

Next, in the expansion of $$\left(\alpha x - \frac{1}{\beta x^3}\right)^{11}$$, the general term is:
$$T_{r+1} = \binom{11}{r}(\alpha x)^{11-r}\left(\frac{-1}{\beta x^3}\right)^r = \binom{11}{r}\frac{(-1)^r\alpha^{11-r}}{\beta^r} x^{11-4r}.$$
Requiring $$11 - 4r = -9$$ yields $$r = 5$$, and hence the coefficient of $$x^{-9}$$ is $$\binom{11}{5}\frac{(-1)^5\alpha^6}{\beta^5} = -\binom{11}{5}\frac{\alpha^6}{\beta^5}\,.$$

From the above, equating these coefficients and noting that $$\binom{11}{6} = \binom{11}{5} = 462$$ gives
$$\frac{\alpha^5}{\beta^6} = -\frac{\alpha^6}{\beta^5}\,,\quad\text{so}\quad \frac{1}{\beta} = -\alpha\,,\quad\text{and thus}\quad \alpha\beta = -1\,. $$

Therefore,
$$(\alpha\beta)^2 = (-1)^2 = 1\,, $$ so the correct answer is 1.