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JEE Main 2026 · 5 Apr · Shift 1, Mathematics question 1

Let a,b∈Ca,b\in\mathbb C. Let α,β\alpha,\beta be the roots of the equation x2+ax+b=0x^2+ax+b=0. If β−α=11\beta-\alpha=\sqrt{11} and β2−α2=3i11\beta^2-\alpha^2=3i\sqrt{11}, then (β3−α3)2\left(\beta^3-\alpha^3\right)^2 is equal to:

Choose one option
Answer and worked solution

Official answer: B

Sum and product. β2−α2=(β−α)(β+α)\beta^2-\alpha^2=(\beta-\alpha)(\beta+\alpha), so α+β=3i1111=3i\alpha+\beta=\dfrac{3i\sqrt{11}}{\sqrt{11}}=3i.
αβ=(α+β)2−(β−α)24=−9−114=−5.\alpha\beta=\frac{(\alpha+\beta)^2-(\beta-\alpha)^2}{4}=\frac{-9-11}4=-5.

Cube difference.
β3−α3=(β−α)[(α+β)2−αβ]=11 (−9+5)=−411.\beta^3-\alpha^3=(\beta-\alpha)\left[(\alpha+\beta)^2-\alpha\beta\right]=\sqrt{11}\,(-9+5)=-4\sqrt{11}.

Square: 16⋅11=17616\cdot11=176.

(B)

Topics:Roots and coefficientsAlgebra of complex numbers
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