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If omega ne 1 is a cube root of unity an...

If `omega ne 1` is a cube root of unity and `a+b=21`, `a^(3)+b^(3)=105`, then the value of `(aomega^(2)+bomega)(aomega+bomega^(2))` is be equal to

A

`3`

B

`5`

C

`7`

D

`35`

Text Solution

Verified by Experts

The correct Answer is:
B

`(b)` `(aomega^(2)+bomega)(aomega+bomega^(2))=a^(2)-ab+b^(2)`
`= ((a+b)(a^(2)-ab+b^(2)))/(a+b)`
`=(a^(3)+b^(3))/(a+b)=(105)/(21)=5`
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