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  4. If a, b, x, y ∈ R , ω ≠ 1, is a cube root of unity and (a+b ω)7=x+y ω, then (b+a ω)7 equals

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Q. If $a, b, x, y \in R , \omega \neq 1$, is a cube root of unity and $(a+b \omega)^7=x+y \omega$, then $(b+a \omega)^7$ equals

Complex Numbers and Quadratic Equations

Solution:

Solution: Taking conjugate, we get
$\left(a+b \omega^2\right)^7 =x+y \omega^2$
$ \Rightarrow\left(a \omega^3+b \omega^2\right)^7=x \omega^3+y \omega^2 $
$\omega^{14}(a \omega+b)^7 =\omega^2(x \omega+y) $
$\Rightarrow (a \omega+b)^7 =x \omega+y$