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Question

If w is a root of the equation x2+x+1=0. The expression
An=nr=1(rw)(rw2) and Bn=nr=1(r+w)(r+w2), then the value of sin[(AnBn)πn] is

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Solution

x2+x+1=0x=1±i32
w=1+i32 which is a cube root of unity.
1+w+w2=0 and w3=1

An=nr=1(rw)(rw2) =nr=1(r2(w+w2)r+w3) =nr=1(r2+r+1)

Bn=nr=1(r+w)(r+w2) =nr=1(r2+(w+w2)r+w3) =nr=1(r2r+1)

AnBn=nr=12r=(n)(n+1)
Hence,
sin[(AnBn)πn]=sin(n+1)π=0

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