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

(C) $-1,1-2 \omega, 1-2 \omega^{2}$

(A) $-1,-1+2 \omega,-1-2 \omega^{2}$

(D) $-1,1+2 \omega, 1+2 \omega^{2}$

JEE Main · 2005 · 08 May 2005 · Offline · Complex Numbers

If the cube roots of unity are

1, ω, ω^{2}

then the roots of the equation

(x - 1)^{3} + 8 = 0

, are

Answer: (C)

Writing the equation as

(x - 1)^{3} = - 8

, we need the three cube roots of

- 8

. One real cube root is

- 2

, and the complete set is obtained by multiplying this by the three cube roots of unity, giving

- 2

- 2 ω

and

- 2 ω^{2}

. Hence

x - 1

equals

- 2

- 2 ω

- 2 ω^{2}

, and the three roots are

x = - 1

x = 1 - 2 ω

and

x = 1 - 2 ω^{2}

JEE Main · 2005 · 08 May 2005 · Offline · Single correct option