If x=a+b,y=aα+bβandz=aβ+bα, where α,β not equal 1are cube roots of unity, then xyz equals
Find the value of xyz:
Given,
x=a+b,y=aα+bβandz=aβ+bα, and α,β are cube roots of unity.
Then,
α=ω,β=ω2
Now,
xyz=(a+b)(aα+bβ)(aβ+bα)=(a+b)a2αβ+abα2+abβ2+b2αβ=(a+b)a2αβ+ab(α2+β2)+b2αβ
Now substitute values of α,β
So,
xyz=(a+b)(a2(ωω2)+ab(ω2+ω4)+b2(ωω2))=(a+b)(a2-ab+b2)[∵ω3=1]=a3+b3
Hence, the correct answer is a3+b3.