If the cube roots of unity are 1,ω,ω², then the roots of (x-1)³+8=0 are (a) -1, 1+2ω, 1+2ω² (b) -1, 1-2ω, 1-2ω² (c) -1,-1,-1 (d) none of these

Correct answer: (b) -1, 1-2ω, 1-2ω² (x-1)³=-8→ x-1=-2, ,-2ω, ,-2ω²→ x=-1, 1-2ω, 1-2ω². JEE Advanced 1979 · Complex Numbers — verified solution by the Vidaara Team.

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[JEE Advanced 1979] if the cube roots of unity are 1 omega omega 2 then the roots

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If the cube roots of unity are $1,\omega,\omega^2$, then the roots of $(x-1)^3+8=0$ are

(a) $-1,\,1+2\omega,\,1+2\omega^2$
(b) $-1,\,1-2\omega,\,1-2\omega^2$
(c) $-1,-1,-1$
(d) none of these

1 Answer

VAVidaara Admin ✓ Vidaara Team ✓ Accepted · 2d ago ▲ 0

Correct answer: (b) $-1,\,1-2\omega,\,1-2\omega^2$

$(x-1)^3=-8\Rightarrow x-1=-2,\,-2\omega,\,-2\omega^2\Rightarrow x=-1,\,1-2\omega,\,1-2\omega^2.$

JEE Advanced 1979 · Complex Numbers — verified solution by the Vidaara Team.

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