Complex NumbersmediumPYQ · JEE Main · 5 Apr 2026 · Shift 1 (Morning)Free

(β³-α³)² from Root Conditions: 176 | JEE Main 2026

JEE Maths question with a full step-by-step solution.

Question
Let a,b∈Ca,b\in\mathbb{C}. Let α,β\alpha,\beta be the roots of 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(\beta^3-\alpha^3)^2 is equal to
A160160
B176176correct
C194194
D187187
Solution
Step 1: β2−α2=(β+α)(β−α)⇒(β+α)11=3i11\beta^2-\alpha^2=(\beta+\alpha)(\beta-\alpha)\Rightarrow(\beta+\alpha)\sqrt{11}=3i\sqrt{11}.
∴ β+α=3i1111=3i.\therefore\ \beta+\alpha=\frac{3i\sqrt{11}}{\sqrt{11}}=3i.
 Step 2: αβ=(β+α)2−(β−α)24=(3i)2−(11)24=−9−114=−204=−5\alpha\beta=\dfrac{(\beta+\alpha)^2-(\beta-\alpha)^2}{4}=\dfrac{(3i)^2-(\sqrt{11})^2}{4}=\dfrac{-9-11}{4}=\dfrac{-20}{4}=-5. Step 3: β3−α3=(β−α)(β2+αβ+α2)=(β−α)[(α+β)2−αβ]\beta^3-\alpha^3=(\beta-\alpha)(\beta^2+\alpha\beta+\alpha^2)=(\beta-\alpha)\big[(\alpha+\beta)^2-\alpha\beta\big].
⇒(β3−α3)2=(β−α)2[(β+α)2−αβ]2.\Rightarrow(\beta^3-\alpha^3)^2=(\beta-\alpha)^2\big[(\beta+\alpha)^2-\alpha\beta\big]^2.
 Step 4: (β−α)2=11, (β+α)2=(3i)2=−9, αβ=−5(\beta-\alpha)^2=11,\ (\beta+\alpha)^2=(3i)^2=-9,\ \alpha\beta=-5:
(β3−α3)2=11[−9−(−5)]2=11(−4)2=11⋅16=176.(\beta^3-\alpha^3)^2=11\big[-9-(-5)\big]^2=11(-4)^2=11\cdot16=176.
 Correct answer: (2)
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