Let $$a, b \in \mathbb{C}$$. Let $$\alpha, \beta$$ be the roots of $$x^2 + ax + b = 0$$. If $$\beta - \alpha = \sqrt{11}$$ and $$\beta^2 - \alpha^2 = 3i\sqrt{11}$$, then $$(\beta^3 - \alpha^3)^2$$ is equal to :

To find the value of $$(\beta^3 - \alpha^3)^2$$, we will use the given equations for the roots $$\alpha$$ and $$\beta$$ of the quadratic equation.

We are given:

1. $$\beta - \alpha = \sqrt{11}$$
2. $$\beta^2 - \alpha^2 = 3i\sqrt{11}$$

Using the difference of squares identity, $$\beta^2 - \alpha^2 = (\beta - \alpha)(\beta + \alpha)$$, we can find the sum of the roots:

$$(\sqrt{11})(\beta + \alpha) = 3i\sqrt{11}$$

$$\beta + \alpha = 3i$$

We can find the individual values of $$\alpha$$ and $$\beta$$ by solving the system:

• $$\beta + \alpha = 3i$$
• $$\beta - \alpha = \sqrt{11}$$

Adding the two equations:

$$2\beta = 3i + \sqrt{11} \implies \beta = \frac{\sqrt{11} + 3i}{2}$$

Subtracting the equations:

$$2\alpha = 3i - \sqrt{11} \implies \alpha = \frac{-\sqrt{11} + 3i}{2}$$

Now, calculate the product $$\alpha\beta$$:

$$\alpha\beta = \left(\frac{3i - \sqrt{11}}{2}\right)\left(\frac{3i + \sqrt{11}}{2}\right) = \frac{(3i)^2 - (\sqrt{11})^2}{4}$$

$$\alpha\beta = \frac{-9 - 11}{4} = \frac{-20}{4} = -5$$

We need to evaluate $$(\beta^3 - \alpha^3)^2$$. First, use the identity for the difference of cubes:

$$\beta^3 - \alpha^3 = (\beta - \alpha)(\beta^2 + \alpha\beta + \alpha^2)$$

Recall that $$\beta^2 + \alpha^2 = (\beta + \alpha)^2 - 2\alpha\beta$$. Therefore:

$$\beta^2 + \alpha\beta + \alpha^2 = (\beta + \alpha)^2 - \alpha\beta$$

Substitute the known values $$(\beta + \alpha = 3i$$ and $$\alpha\beta = -5)$$:

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

Now, substitute this back into the difference of cubes:

$$\beta^3 - \alpha^3 = (\sqrt{11})(-4) = -4\sqrt{11}$$

Finally, square the result:

$$(\beta^3 - \alpha^3)^2 = (-4\sqrt{11})^2$$

$$= 16 \times 11$$

$$= 176$$

Final Answer: 176 (Option B)