We are given the roots $\alpha$ and $\beta$ of the quadratic equation $x^2 + ax + b = 0$. We are provided with two conditions:
- $\beta - \alpha = \sqrt{11}$
- $\beta^2 - \alpha^2 = 3i\sqrt{11}$
We need to find the value of $(\beta^3 - \alpha^3)^2$.
Using Given Conditions
Factor the second condition:
$\beta^2 - \alpha^2 = (\beta - \alpha)(\beta + \alpha)$
Substitute the given values:
$3i\sqrt{11} = (\sqrt{11})(\beta + \alpha)$
Divide both sides by $\sqrt{11}$ to find the sum of the roots:
$\beta + \alpha = 3i$
Calculating $\beta^3 - \alpha^3$
Recall the identity for the difference of cubes:
$\beta^3 - \alpha^3 = (\beta - \alpha)(\beta^2 + \alpha\beta + \alpha^2)$
We can rewrite $\beta^2 + \alpha\beta + \alpha^2$ using the sum and product of roots. We know:
$\beta^2 + \alpha\beta + \alpha^2 = (\beta + \alpha)^2 - \alpha\beta$
From Vieta's formulas for the equation $x^2 + ax + b = 0$, we have:
- $\alpha + \beta = -a$
- $\alpha\beta = b$
We already found $\beta + \alpha = 3i$, so $-a = 3i$, which means $a = -3i$.
To find $\alpha\beta$, we can solve for $\alpha$ and $\beta$ using the two sum/difference equations:
- Adding $(\beta - \alpha = \sqrt{11})$ and $(\beta + \alpha = 3i)$ gives $2\beta = \sqrt{11} + 3i$, so $\beta = \frac{\sqrt{11} + 3i}{2}$.
- Subtracting $(\beta - \alpha = \sqrt{11})$ from $(\beta + \alpha = 3i)$ gives $2\alpha = 3i - \sqrt{11}$, so $\alpha = \frac{-\sqrt{11} + 3i}{2}$.
Now, calculate the product $\alpha\beta$:
$\alpha\beta = \left( \frac{-\sqrt{11} + 3i}{2} \right) \left( \frac{\sqrt{11} + 3i}{2} \right) = \frac{(3i)^2 - (\sqrt{11})^2}{4} = \frac{-9 - 11}{4} = \frac{-20}{4} = -5$
So, $b = -5$.
Now substitute the values into the expression for $\beta^2 + \alpha\beta + \alpha^2$:
$\beta^2 + \alpha\beta + \alpha^2 = (\beta + \alpha)^2 - \alpha\beta = (3i)^2 - (-5) = -9 + 5 = -4$
Substitute this back into the formula for $\beta^3 - \alpha^3$:
$\beta^3 - \alpha^3 = (\beta - \alpha)(\beta^2 + \alpha\beta + \alpha^2) = (\sqrt{11})(-4) = -4\sqrt{11}$
Final Calculation
Finally, calculate $(\beta^3 - \alpha^3)^2$:
$(\beta^3 - \alpha^3)^2 = (-4\sqrt{11})^2 = (-4)^2 \times (\sqrt{11})^2 = 16 \times 11 = 176$
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