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If $\omega(\neq1)$ is a cube root of unity and $(1+\omega)^7=A+B\omega$, then $(A,B)=?$

$\begin{array}{1 1}(A) \;(0,1) \\(B)\;(1,1) \\(C)\;(1,0)\\(D)\;(-1,1) \end{array}$

1 Answer

Toolbox:
  • $1+\omega+\omega^2=0$
  • $\omega^n=1$ if $ n$ is divisible by 3
  • $=\omega$, if $n$ leaves remainder 1 when divided by 3
  • $=\omega^2$, if $n$ leaves remainder 2 when divided by 3
$1+\omega+\omega^2=0$
$\Rightarrow\:(1+\omega)=-\omega^2$
$\Rightarrow(1+\omega)^7=(-\omega^2)^7$
$=-\omega^{14}$
$=-\omega^2$  (Because 14 leaves remainder 2 when divided by 3)
$=1+\omega$
answered Jul 3, 2013 by rvidyagovindarajan_1
edited May 20, 2014 by rohanmaheshwari0831_1
 

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