Question

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

• A. $y+x\omega$
• B. $-y-x\omega$
• C. $x+y\omega$
• D. $-x-y\omega$

Answer 1

Answer: (A)

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