(x³ - 1) Where ω is one of the cube roots of unity.

Correct Answer is : (x−1)(x−ω)(x−ω2)(x - 1) (x - ω) (x - ω²)(x−1)(x−ω)(x−ω2)

Correct Answer is : (x−1)(x−ω)(x−ω2)(x - 1) (x - ω) (x - ω²)(x−1)(x−ω)(x−ω2)

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UPSC NDA I 2015 Mathematics Paper

Question : 21 of 120

Marks: +1, -0

(x^{3} - 1)

Where ω is one of the cube roots of unity.

Solution:

Concept:

Use the identity

x^3 - y^3 = (x - y)(x^2 + xy + y^2)

and express

x^2 + x + 1

using cube roots of unity:

\omega

and

\omega^2

, where

\omega^3 = 1

and

1 + \omega + \omega^2 = 0

Explanation:

Start with

x^3 - 1 = (x - 1)(x^2 + x + 1)

Now factor

x^2 + x + 1

. Solve

x^2 + x + 1 = 0

x = \frac{-1 \pm \sqrt{1 - 4}}{2} = \frac{-1 \pm i\sqrt{3}}{2}

These two roots are exactly

\omega

and

\omega^2

, the complex cube roots of unity.

Thus

x^2 + x + 1 = (x - \omega)(x - \omega^2)

Substitute back:

x^3 - 1 = (x - 1)(x - \omega)(x - \omega^2)

Answer:

Option B:

(x - 1)(x - \omega)(x - \omega^2)

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