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Question

If α & β are complex cube rout of unity x=a+b ,y=αa+βb ,z=aβ+bα then show xyz=a3+b3

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Solution

Consider the problem

Let,

We denote complex cube root of unity as r and r2

where
1+r+r2=0

Also,

r3=1

Now,

α=r and β=r2
x=a+b

y=aα+bβ=ar+br2z=aβ+bα=ar2+brxyz=(a+b)(ar+br2)(ar2+br)=(a+b)(a2r3+abr2+abr4+b2r3)=(a+b)(a2r3+abr2+abr3.r+b2r3)

As r3=1

Then,

xyz=(a+b)(a2+abr2+abr+b2)=(a+b)(a2+ab(r2+r)+b2)

As 1+r+r2=0

r+r2=1

Then,

=(a+b)(a2ab+b2)xyz=a3b3

Hence, proved.

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