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

Question

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

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`

Monjur Alahi

5 views

2025-01-04 04:42

1 Answers

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Salim Ahmed Kawsar · Accepted Answer · 2023-02-05T15:29:48+06:00

Salim Ahmed Kawsar

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`

5 views Answered 2023-02-05 15:29