1please explain how the answer is arrieved at .
If α, β,γ are the cube roots of p,(p<0), then for any x,y,z (α*x+β*y+γ*z)/(β*x+γ*y+α*z) is equal to
(A) α *ω+β*ω2+γ
(B) α*β*γ
(c) ω,ω2
(d) none
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8 Answers
11rashi u can use the symbols in GREEKS please edit this question to make it clear[1]
62\frac{(\alpha x+\beta y+\gamma z)}{(\beta x+\gamma y+\alpha z)}
Its answer should be either w or w2
62see \alpha \omega = \beta or \alpha \omega = \gamma
Take 1st case, then \beta \omega = \gamma
If you can figure this much out, you are done!
1the cube roots of p are p1/3ω , p1/3ω2 and p1/3
α =p1/3ω , β= p1/3ω2 , γ=p1/3
then
(α*x+β*y+γ*z)/(β*x+γ*y+α*z) = (p1/3ω*x+p1/3ω2*y+p1/3*z)/(p1/3ω2*x+p1/3*y+p1/3ω*z)
multiply and divide the numerator and denominator by ω
take ω inside the bracket of numerator then the numerator will be identical to the term in the bracket of the denominator .. cancel both
u will get 1/ω which is w2
interchange the values of α & β then u will get ω as answer
1first how the consideration of ω is taken as cube root of unity is not involved in the question , if involved how?, where is p ? how the roots are substituted please explain in detail.
1if u will consider the complex cube root of p is (p.1)1/3 = (p)1/3.(1)1/3 = (p)1/3ω , (p)1/3ω2 & of course (p)1/3