Question

If `1/(a + ω) + 1/(b + ω) + 1/(c + ω) + 1/(d + ω) = 1/ω`, where a, b, c, d ∈ R and ω is a cube root of unity then `sum 3/(a^2 - a + 1)` is equal to

पर्याय

• 1

• 2

• 3

• `1/3`

Answer 1

3

Explanation:

Given that `1/(a + ω) + 1/(b + ω) + 1/(c + ω) + 1/(d + ω) = 1/w`  ......(i)

Taking conjugate we get,

`1/(a + ω^2) + 1/(b + ω^2) + 1/(c + ω^2) + 1/(d + ω^2) + 1/ω^2`  .....(ii)

Subtract equation (ii) from equation (i) we get

`sum(1/(a + ω) 1/(a + ω^2)) = 1/ω - 1/ω^2`

⇒ `sum 1/((a + ω)(a + ω^2)) = ω^2 - ω`

So `sum 1/(a^2 - a + 1)` = 1

⇒ `sum 3/(a^2 - a + 1)` = 3