Question
Let α, β and γ be the cube roots of p (p < 0). For any real numbers x, y and z, the value of (xα+yβ+zγ) ⁄(xβ+yγ+zα) is
According to the given condition p is a negative number
consider p = -k so cube root of p will be α = -k1/3, β = -k1/3ω, γ = -k1/3ω2
On applying given values you will get the answer as ω or ω2
you should use ω3 = 1
According to the given condition p is a negative number
consider p = -k = (k1/3)3(-1) so cube root of p will be α = -k1/3, β = -k1/3ω, γ = -k1/3ω2 , here -1, -ω, -ω2 are the cube roots of -1
On applying given values on the given expression, you will get
x(-k1/3)+y(-k1/3ω)+z(-k1/3ω2)/x(-k1/3ω)+y(-k1/3ω2)+z(-k1/3)
=x+yω+zω2/xω+yω2+z
first multiply ω in numerator and denominator we get answer as 1/ω = ω2
second interchange the value of α, β, γ for getting answer as ω
you should use ω3 = 1
Read 3 Solution.
lets say p=-1 then its root are -1,-ω,-ω2
so the above eqn has roots p,pω,pω2
now divide and multiply the above equation by ω after that u will find that the above equation get reduced to the form 1/ω .
therefore the answer is -ω
i don't get the answer. sir, can u please solve it.
use p1/3instead of p
1/w=w3/w=w2