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If $\omega$ is a cube root of unity then the value of $(1-\omega+\omega^{2})^{4}+(1+\omega-\omega^{2})^{4}$ is

\[\begin{array}{1 1}(1)0&(2)32\\(3)-16&(4)-32\end{array}\]

1 Answer

$\omega$ is a cube root of unity
$\omega^3=1$ and $1+\omega+\omega^2=0$
$(1-\omega+\omega^2)^4+(1+\omega-\omega^2)^4 =(1+ \omega^2 - \omega)^4+ (1+\omega- \omega^2)^4$
$\qquad= (- \omega - \omega)^4+(- \omega^2-\omega^2)^4$
$\qquad=(-2 \omega)^4+(-2 \omega^2)^4$
$\qquad= 2^4\omega^4+2^4 \omega^8$
$\qquad= 2^4(\omega^{4}+ \omega^{8})$
$\qquad= 16 (\omega^3. \omega + \omega^3. \omega^3. \omega^2)$
$\qquad= 16 (\omega +\omega^2)$
$\qquad= 16 \times -1$
$\qquad=-16$
Hence 3 is the correct answer.
answered May 15, 2014 by meena.p
 
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