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

- 1, - 1 + 2 w, - 1 - 2 w^{2}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

-1 , -1, - 1

No worries! We‘ve got your back. Try BYJU‘S free classes today!

- 1, 1 - 2 w, 1 - 2 w^{2}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

- 1, 1 + 2 w, 1 + 2 w^{2}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is D $- 1, 1 - 2 w, 1 - 2 w^{2}$
The roots of the equation $(x - 1)^{3} + 8 = 0 \Rightarrow (x - 1) = (- 2) (1)^{1 / 3}$
$\Rightarrow x - 1 = - 2$ or $- 2 w^{2}$ or $- 2 ω$
$\Rightarrow x = - 1$ or $1 - 2 w^{2} .$ or $1 - 2 ω$

Suggest Corrections

Similar questions

w

a, b, c

\frac{1}{a + w} + \frac{1}{b + w} + \frac{1}{c + w} = 2 w^{2}

\frac{1}{a + w^{2}} + \frac{1}{b + w^{2}} + \frac{1}{c + w^{2}} = 2 w

\frac{1}{a + 1} + \frac{1}{b + 1} + \frac{1}{c + 1}

1, w, w, w^{2}

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

1, ω, ω^{2}

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

1, w, w^{2}

2 w, (2 + 3 w), (2 + 3 w^{2}), (2 - w - w^{2})

1, w, w^{2}

(1 - w + w^{2}) (1 + w - w^{2})

Join BYJU'S Learning Program