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State true or false.

The sum of cube roots of unity is zero.

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True

False

answer is A.

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Detailed Solution

Cube root of unity is the numbers which is when raised by the power of three, generates the number one.
The cube root of unity can be given as,

ω = \sqrt[3]{1}

Take cube on both sides,

ω^{3} = 1

ω^{3} - 1 = 0 \dots \dots \dots (1)

ω^{3} - 1^{3} = (ω - 1) (ω^{2} + ω + 1^{2})

(since,

a^{3} - b^{3} = (a - b) (a^{2} + ab + b^{2})

)

ω^{3} - 1 = (ω - 1) (ω^{2} + ω + 1)

……. (2)
From equation (1) we get,

(ω - 1) (ω^{2} + ω + 1) = 0

ω - 1 = 0; ω^{2} + ω + 1 = 0

Here, one solution,

ω = 1

Now solve the quadratic equation,

ω^{2} + ω + 1 = 0

On comparing with

a x^{2} + bx + c = 0

the above equation, put the values in following quadratic formula,

x = \frac{- b \pm \sqrt{b^{2} - 4 ac}}{2 a}

ω = \frac{- 1 \pm \sqrt{1^{2} - 4 (1) (1)}}{2 (1)}

ω = \frac{- 1 \pm \sqrt{1 - 4}}{2}

ω = \frac{- 1 \pm \sqrt{- 3}}{2}

Since,

\sqrt{- 3}

i \sqrt{3}

. So,

ω = \frac{- 1 \pm i \sqrt{3}}{2}

So, cube roots of unity are

\frac{- 1 + i \sqrt{3}}{2}, \frac{- 1 - i \sqrt{3}}{2}

and 1.
Sum of cube root of unity,

\frac{- 1 + i \sqrt{3}}{2} + \frac{- 1 - i \sqrt{3}}{2} + 1 = - \frac{1}{2} - \frac{1}{2} + 1

\frac{- 1 + i \sqrt{3}}{2} + \frac{- 1 - i \sqrt{3}}{2} + 1 = - 1 + 1

\frac{- 1 + i \sqrt{3}}{2} + \frac{- 1 - i \sqrt{3}}{2} + 1 = 0

So, the given statement is true.
Option 1 is Correct.

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