JEE Main 2026 — Mathematics Algebra

Given $\beta - \alpha = \sqrt{11}$β−α=11​ and $\beta^2 - \alpha^2 = 3i\sqrt{11}$β2−α2=3i11​

$$\beta + \alpha = \dfrac{\beta^2 - \alpha^2}{\beta - \alpha} = \dfrac{3i\sqrt{11}}{\sqrt{11}} = 3i$$β+α=β−αβ2−α2​=11​3i11​​=3i

Using the identity $(\beta - \alpha)^2 = (\beta + \alpha)^2 - 4\alpha\beta$(β−α)2=(β+α)2−4αβ

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

$$11 = -9 - 4\alpha\beta$$11=−9−4αβ

$$4\alpha\beta = -20 \Rightarrow \alpha\beta = -5$$4αβ=−20⇒αβ=−5

Now, $\beta^3 - \alpha^3 = (\beta - \alpha)(\beta^2 + \alpha\beta + \alpha^2)$β3−α3=(β−α)(β2+αβ+α2)

$$\beta^3 - \alpha^3 = (\beta - \alpha)((\beta + \alpha)^2 - \alpha\beta)$$β3−α3=(β−α)((β+α)2−αβ)

$$\beta^3 - \alpha^3 = \sqrt{11}((3i)^2 - (-5))$$β3−α3=11​((3i)2−(−5))

$$\beta^3 - \alpha^3 = \sqrt{11}(-9 + 5) = -4\sqrt{11}$$β3−α3=11​(−9+5)=−411​

Squaring both sides, we get:

$$(\beta^3 - \alpha^3)^2 = (-4\sqrt{11})^2 = 16 \times 11 = 176$$(β3−α3)2=(−411​)2=16×11=176

Answer: $176$176