If $$\alpha$$ and $$\beta$$ are the distinct roots of the equation $$x^2 + (3)^{1/4}x + 3^{1/2} = 0$$, then the value of $$\alpha^{96}(\alpha^{12} - 1) + \beta^{96}(\beta^{12} - 1)$$ is equal to:

The equation is $$x^2 + 3^{1/4}x + 3^{1/2} = 0$$. Let $$\omega = 3^{1/4}$$, so the equation becomes $$x^2 + \omega x + \omega^2 = 0$$.

The roots are: $$x = \frac{-\omega \pm \sqrt{\omega^2 - 4\omega^2}}{2} = \frac{-\omega \pm \sqrt{-3\omega^2}}{2} = \frac{-\omega \pm i\omega\sqrt{3}}{2} = \omega \cdot \frac{-1 \pm i\sqrt{3}}{2}$$

Recognizing that $$\frac{-1 \pm i\sqrt{3}}{2}$$ are the complex cube roots of unity, we have $$\alpha = \omega \cdot \zeta$$ and $$\beta = \omega \cdot \zeta^2$$, where $$\zeta = e^{2\pi i/3}$$ satisfies $$\zeta^3 = 1$$ and $$1 + \zeta + \zeta^2 = 0$$.

Now compute $$\alpha^{96}(\alpha^{12} - 1) + \beta^{96}(\beta^{12} - 1)$$.

First, $$\alpha^{12} = \omega^{12} \cdot \zeta^{12} = (3^{1/4})^{12} \cdot (\zeta^3)^4 = 3^3 \cdot 1 = 27$$. Similarly, $$\beta^{12} = \omega^{12} \cdot \zeta^{24} = 27 \cdot (\zeta^3)^8 = 27$$.

So $$\alpha^{12} - 1 = 26$$ and $$\beta^{12} - 1 = 26$$.

Next, $$\alpha^{96} = \omega^{96} \cdot \zeta^{96} = (3^{1/4})^{96} \cdot (\zeta^3)^{32} = 3^{24} \cdot 1 = 3^{24}$$. Similarly, $$\beta^{96} = 3^{24}$$.

Therefore: $$\alpha^{96}(\alpha^{12}-1) + \beta^{96}(\beta^{12}-1) = 3^{24} \cdot 26 + 3^{24} \cdot 26 = 2 \times 26 \times 3^{24} = 52 \times 3^{24}$$

This corresponds to Option 3: $$52 \times 3^{24}$$.