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Value(s) of ${(- i)}^{\frac{1}{3}}$ is/are:

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\frac{\sqrt{3} - i}{2}

\frac{\sqrt{3} + i}{2}

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

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answer is A.

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Detailed Solution

We can write

{(- i)}^{\frac{1}{3}}

=z.
Cubing both the sides, we will get:
⇒ −i=

z^{3}

Using the value −i=

i^{3}

and bringing all the terms to the same side of the equation, we get:
⇒

z^{3} - i^{3}

On using the factorization

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

, we have:
⇒ (z−i)(

z^{2}

+zi+

i^{2}

)=0
Using

i^{2}

=−1 and the fact that the product of two terms can be 0 only if at least one of them is 0, we conclude the following:
⇒ z−i=0 OR

z^{2}

+iz−1=0
Using the quadratic formula, we get:
⇒ z=i OR z=

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

⇒ z=i OR z=

\frac{- i \pm \sqrt{- 1 + 4}}{2}

⇒ z=i OR z= =

\frac{- i \pm \sqrt{3}}{2}

OR
⇒ z= =

\frac{- i - \sqrt{3}}{2}

∴ The correct answer options are

\frac{\sqrt{3 - i}}{2}

and

\frac{- \sqrt{3} - i}{2}

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