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 toA. `3`B. `5`C. `7`D. `35`

If `omega ne 1` is a cube root of unity and `a+b=21`, `a^(3)+b^(3)=105

1.	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 toA. `3`B. `5`C. `7`D. `35`
Answer» Correct Answer - 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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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 toA. `3`B. `5`C. `7`D. `35`

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