Question

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

• A. 160
• B. 176
• C. 194
• D. 187

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
We are given a quadratic equation $x^2 + ax + b = 0$ with complex coefficients and roots $\alpha, \beta$.
We are provided with two relations involving the roots: $\beta - \alpha = \sqrt{11}$ and $\beta^2 - \alpha^2 = 3i\sqrt{11}$.
Our goal is to find the value of $(\beta^3 - \alpha^3)^2$.
Step 2: Key Formula or Approach:
We will use the given relations to find the sum of the roots ($\alpha + \beta$) and the product of the roots ($\alpha\beta$).
The key identities are:
1. $\beta^2 - \alpha^2 = (\beta - \alpha)(\beta + \alpha)$
2. $\beta^3 - \alpha^3 = (\beta - \alpha)(\beta^2 + \alpha\beta + \alpha^2)$
3. $\beta^2 + \alpha^2 = (\beta + \alpha)^2 - 2\alpha\beta$
Combining (2) and (3), we get $\beta^3 - \alpha^3 = (\beta - \alpha)((\beta + \alpha)^2 - \alpha\beta)$.
Step 3: Detailed Explanation:
We are given:
\[ \beta - \alpha = \sqrt{11} \quad \text{---(1)} \] \[ \beta^2 - \alpha^2 = 3i\sqrt{11} \quad \text{---(2)} \] Using the identity $\beta^2 - \alpha^2 = (\beta - \alpha)(\beta + \alpha)$, we can substitute the given values:
\[ 3i\sqrt{11} = (\sqrt{11})(\beta + \alpha) \] Dividing by $\sqrt{11}$ on both sides, we get:
\[ \beta + \alpha = 3i \quad \text{---(3)} \] Now we find the product of the roots, $\alpha\beta$. We know that for the quadratic equation $x^2+ax+b=0$, the product of roots is $b$.
We use the identity $(\beta - \alpha)^2 = (\beta + \alpha)^2 - 4\alpha\beta$.
Substituting the values from (1) and (3):
\[ (\sqrt{11})^2 = (3i)^2 - 4\alpha\beta \] \[ 11 = 9i^2 - 4\alpha\beta \] Since $i^2 = -1$:
\[ 11 = -9 - 4\alpha\beta \] \[ 20 = -4\alpha\beta \] \[ \alpha\beta = -5 \] Now, we need to calculate $\beta^3 - \alpha^3$.
Using the identity $\beta^3 - \alpha^3 = (\beta - \alpha)((\beta + \alpha)^2 - \alpha\beta)$:
\[ \beta^3 - \alpha^3 = (\sqrt{11})((3i)^2 - (-5)) \] \[ \beta^3 - \alpha^3 = \sqrt{11}(-9 + 5) \] \[ \beta^3 - \alpha^3 = \sqrt{11}(-4) = -4\sqrt{11} \] Finally, we need to find $(\beta^3 - \alpha^3)^2$:
\[ (\beta^3 - \alpha^3)^2 = (-4\sqrt{11})^2 \] \[ (\beta^3 - \alpha^3)^2 = (-4)^2 \times (\sqrt{11})^2 = 16 \times 11 = 176 \] Step 4: Final Answer:
The value of $(\beta^3 - \alpha^3)^2$ is 176.