Q26 / 75 · JEE Main 2026 (05 April Shift 1) · Mathematics

Let $a, b \in \mathbb{C}$. 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
    $160$
  • Correct answer
    B
    $176$
  • C
    $194$
  • D
    $187$

Solution

Given $\beta - \alpha = \sqrt{11}$ and $\beta^2 - \alpha^2 = 3i\sqrt{11}$

$\beta + \alpha = \dfrac{\beta^2 - \alpha^2}{\beta - \alpha} = \dfrac{3i\sqrt{11}}{\sqrt{11}} = 3i$

Using the identity $(\beta - \alpha)^2 = (\beta + \alpha)^2 - 4\alpha\beta$

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

$11 = -9 - 4\alpha\beta$

$4\alpha\beta = -20 \Rightarrow \alpha\beta = -5$

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

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

$\beta^3 - \alpha^3 = \sqrt{11}((3i)^2 - (-5))$

$\beta^3 - \alpha^3 = \sqrt{11}(-9 + 5) = -4\sqrt{11}$

Squaring both sides, we get:

$(\beta^3 - \alpha^3)^2 = (-4\sqrt{11})^2 = 16 \times 11 = 176$

Answer: $176$

Hence, the correct answer is option: 'B'.

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