If the range of $$f(\theta) = \frac{\sin^4\theta + 3\cos^2\theta}{\sin^4\theta + \cos^2\theta}, \theta \in \mathbb{R}$$ is $$[\alpha, \beta]$$, then the sum of the infinite G.P., whose first term is 64 and the common ratio is $$\frac{\alpha}{\beta}$$, is equal to ________

We set $$t = \cos^2\theta \in [0,1]$$ so that $$\sin^2\theta = 1 - t$$ and $$\sin^4\theta = (1 - t)^2$$, and substituting these into $$f(\theta)$$ yields $$f = \frac{(1-t)^2 + 3t}{(1-t)^2 + t} = \frac{1-2t+t^2+3t}{1-2t+t^2+t} = \frac{t^2+t+1}{t^2-t+1}.$$

Letting $$y = \frac{t^2 + t + 1}{t^2 - t + 1}$$ leads to $$y(t^2 - t + 1) = t^2 + t + 1$$ which rearranges to $$(y-1)t^2 - (y+1)t + (y-1) = 0.$$

Since this quadratic in $$t$$ must have a real root in $$[0,1]$$, its discriminant satisfies $$(y+1)^2 - 4(y-1)^2 \ge 0;$$ noting that $$(y+1)^2 - (2y-2)^2 = (3y-1)(3-y)$$ we obtain $$(3y-1)(3-y) \ge 0,$$ which implies $$\tfrac13 \le y \le 3.$$

Checking the endpoints $$t = 0$$ and $$t = 1$$ gives $$f(0) = 1$$ and $$f(1) = 3$$, and no smaller value occurs for $$t \in [0,1]$$; therefore the range of $$f$$ is $$[1,3]$$, so that $$\alpha = 1$$ and $$\beta = 3$$.

Finally, the sum of the infinite geometric progression with first term $$a = 64$$ and common ratio $$r = \alpha/\beta = 1/3$$ is $$S = \frac{a}{1-r} = \frac{64}{1-1/3} = \frac{64}{2/3} = 96.$$

Hence the final answer is 96.