Let be the roots of the equaiton such that is in the form where are positive integers and as the conjugate of .
Given that . Find the value of .
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Clearly α , β and γ are the cubic roots of unity. We know that α = 1 , β = e 3 2 π i and γ = e − 3 2 π i . So,
x = e 3 2 π i − e − 3 2 π i − 1 − 1 x = 2 i sin ( 3 2 π ) − 2 x = 3 i − 2
Manipulate that expression a little:
x + 2 = 3 i ( x + 2 ) 2 = ( 3 i ) 2 x 2 + 4 x + 4 = − 3 x 2 + 4 x + 7 = 0
Now, using long polynomial division we know that:
x 4 + 3 x 3 + 2 x 2 − 1 1 x − 6 = ( x 2 + 4 x + 7 ) ( x 2 − x − 1 ) + 1 x 4 + 3 x 3 + 2 x 2 − 1 1 x − 6 = ( 0 ) ( x 2 − x − 1 ) + 1 x 4 + 3 x 3 + 2 x 2 − 1 1 x − 6 = 1