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Question

.x3-1 can be factorised as: where ω is one of the cube roots of unity.


A

(x-1)(xω)(x+ω2)

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B

(x-1)(xω)(x-ω2)

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C

(x-1)(x+ω)(x+ω2)

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D

(x-1)(x+ω)(x-ω2)

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Solution

The correct option is B

(x-1)(xω)(x-ω2)


Explanation for the correct option.

We know that

.x3-1=x-1x2+x+1

Now, roots of x2+x+1 are:

x=-1±12-4×1×12×1=-1±1-42=-1±-32=-1±3i2-1=i

Therefore, -1+3i2,-1-3i2 are the roots of x2+x+1.

Now,

ω=-1+i32ω2=-1-i32

Hence 1,ω,ω2 are the roots.

Thus, factors are (x1)(xω) and x-ω2.

Hence, the correct option is B.


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