If α, β, γ are the cube roots of p, p

Question

If α, β, &gamma; are the cube roots of p, p<0, then for any x, y and z which does not make denominator zero, the expression (x α+y β+z &gamma;/x β+y &gamma;+z α) equals (A) ω, 1 (B) ω, ω2 (C) ω2, 1 (D) 1, ω, ω2

Tardigrade Admin · Accepted Answer

x3=p=(p(1/3))3 ⇒ x=p(1/3), p(1/3) ω, p(1/3) ω2 Let α=p(1/3), β=p(1/3) ω, &gamma;=p(1/3) ω2. (x α+y β+z &gamma;/x β+y &gamma;+z α)=(x+y ω+ω2 z/x ω+y ω2+z)=(1/ω)=ω2 If α=p(1/3) ω, β=p(1/3), &gamma;=p(1/3) ω2, then (x α+y β+z &gamma;/x β+y &gamma;+z α)=(1/ω2)=ω

Q. If $α, β, γ$ are the cube roots of $p, p < 0$ , then for any $x, y$ and $z$ which does not make denominator zero, the expression $\frac{xα + y β + z γ}{x β + y γ + z α}$ equals

$ω, 1$

$ω, ω^{2}$

$ω^{2}, 1$

$1, ω, ω^{2}$

Q. If α,β,γ are the cube roots of p,p<0, then for any x,y and z which does not make denominator zero, the expression xβ+yγ+zαxα+yβ+zγ​ equals

ω,1

ω,ω2

ω2,1

1,ω,ω2

x3=p=(p31​)3⇒x=p31​,p31​ω,p31​ω2 Let α=p31​,β=p31​ω,γ=p31​ω2. xβ+yγ+zαxα+yβ+zγ​=xω+yω2+zx+yω+ω2z​=ω1​=ω2 If α=p31​ω,β=p31​,γ=p31​ω2, then xβ+yγ+zαxα+yβ+zγ​=ω21​=ω

Q. If $α, β, γ$ are the cube roots of $p, p < 0$ , then for any $x, y$ and $z$ which does not make denominator zero, the expression $\frac{xα + y β + z γ}{x β + y γ + z α}$ equals

$ω, 1$

$ω, ω^{2}$

$ω^{2}, 1$

$1, ω, ω^{2}$