If α, β,γ are the roots of x3−3x2+3x+7=0 and ω is a complex cube root of unity, then the value of (α−1β−1+β−1γ−1+γ−1α−1) is equal to p(ωq) for some p,q∈Z. Then the number of ordered pairs (p,q) such that p+q=15 is

α,β,γ are roots of (x−1)3=−8⇒α−1, β−1,γ−1 are equal to −2, −2ω, −2ω2 in any order then α−1β−1+β−1γ−1+γ−1α−1=3ω2 or 3ω = 3(ω)3k+1 or 3(ω)3k+2 where k∈Z∴ p= 3, q = 3k+1 or 3k+2p+q=15⇒k=113 or 103 which is not possible

If α, β,γ are the roots of x3−3x2+3x+7=0 and ω is a complex cube root of unity, then the value of (α−1β−1+β−1γ−1+γ−1α−1) is equal to p(ωq) for some p,q∈Z. Then the number of ordered pairs (p,q) such that p+q=15 is

α,β,γ are roots of (x−1)3=−8⇒α−1, β−1,γ−1 are equal to −2, −2ω, −2ω2 in any order then α−1β−1+β−1γ−1+γ−1α−1=3ω2 or 3ω = 3(ω)3k+1 or 3(ω)3k+2 where k∈Z∴ p= 3, q = 3k+1 or 3k+2p+q=15⇒k=113 or 103 which is not possible

AI Mentor

Check Your IQ

Free Expert Demo

Try Test

Courses

Dropper NEET Course Dropper JEE Course Class - 12 NEET Course Class - 12 JEE Course Class - 11 NEET Course Class - 11 JEE Course Class - 10 Foundation NEET Course Class - 10 Foundation JEE Course Class - 10 CBSE Course Class - 9 Foundation NEET Course Class - 9 Foundation JEE Course Class -9 CBSE Course Class - 8 CBSE Course Class - 7 CBSE Course Class - 6 CBSE Course

Offline Centres

If $α, β, γ$ are the roots of $x^{3} - 3 x^{2} + 3 x + 7 = 0$ and $ω$ is a complex cube root of unity, then the value of $(\frac{α - 1}{β - 1} + \frac{β - 1}{γ - 1} + \frac{γ - 1}{α - 1})$ is equal to $p (ω^{q})$ for some $p, q \in Z .$ Then the number of ordered pairs $(p, q)$ such that $p + q = 15$ is

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

Infinite

answer is A.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

$α, β, γ$ are roots of ${(x - 1)}^{3} = - 8$

$\Rightarrow α - 1, β - 1, γ - 1$ are equal to $- 2, - 2 ω, - 2 ω^{2}$ in any order then $\frac{α - 1}{β - 1} + \frac{β - 1}{γ - 1} + \frac{γ - 1}{α - 1} = 3 ω^{2} o r 3 ω = 3 {(ω)}^{3 k + 1} o r 3 {(ω)}^{3 k + 2} w h e r e k \in Z$