Roots of Quadratic Equation with Complex Coefficients

Given that $\alpha$ and $\beta$ are the roots of the quadratic equation $x^2 + ax + b = 0$, we are provided with the following relations: $\beta - \alpha = \sqrt{11}$ $\beta^2 - \alpha^2 = 3i\sqrt{11}$

Since $\beta^2 - \alpha^2 = (\beta - \alpha)(\beta + \alpha)$, we can substitute the value of $\beta - \alpha$: $\sqrt{11}(\beta + \alpha) = 3i\sqrt{11}$ $\implies \beta + \alpha = 3i$

Now, using the algebraic identity for the product of roots $\alpha\beta$, we have: $\alpha\beta = \frac{(\beta + \alpha)^2 - (\beta - \alpha)^2}{4}$ Substituting the known values: $\alpha\beta = \frac{(3i)^2 - (\sqrt{11})^2}{4} = \frac{-9 - 11}{4} = \frac{-20}{4} = -5$

Next, we expand $\beta^3 - \alpha^3$ using the factorization formula: $\beta^3 - \alpha^3 = (\beta - \alpha)(\beta^2 + \alpha\beta + \alpha^2)$

Rewriting $\beta^2 + \alpha\beta + \alpha^2$ in terms of $(\beta + \alpha)$ and $\alpha\beta$: $\beta^2 + \alpha\beta + \alpha^2 = (\beta + \alpha)^2 - \alpha\beta$

Substituting the values: $\beta^2 + \alpha\beta + \alpha^2 = (3i)^2 - (-5) = -9 + 5 = -4$

Thus, $\beta^3 - \alpha^3$ becomes: $\beta^3 - \alpha^3 = \sqrt{11} \times (-4) = -4\sqrt{11}$

Finally, we calculate $(\beta^3 - \alpha^3)^2$: $(\beta^3 - \alpha^3)^2 = (-4\sqrt{11})^2 = 16 \times 11 = 176$

Hence, the correct option is B.