Question

If 1,w,w^2 are cube Roots unity, show that (2-w) (2-w^2) (2-w^10) (2-w^11) =49

Answer 1

$(2 - \omega)(2 - { \omega}^{2} )(2 - { \omega}^{10})(2 - { \omega}^{11}) \\ = (2 - \omega)(2 - { \omega}^{2} )(2 - \omega \times \:{ \omega}^{9} ) (2 - { \omega}^{2} \times { \omega}^{9}) \\ = (2 - \omega)(2 - { \omega}^{2}) (2 - \omega \times 1)(2 - { \omega}^{2} \times 1) - - - - - - (since \: {\omega}^{3} ) = 1 \\ = (2 - \omega)(2 - { \omega}^{2} )(2 - \omega)(2 - { \omega}^{2} ) \\ = (4 - 2 \omega - 2 { \omega}^{2} + { \omega}^{3}) (4 - 2 \omega - 2 { \omega}^{2} + { \omega}^{3}) \\ = (4 - 2( \omega + { \omega}^{2}) + 1) (4 - 2( \omega + { \omega}^{2}) + 1)\\=(4 - 2( - 1) + 1)(4 - 2( - 1) + 1) \\ = (4 + 2 + 1)(4 + 2 + 1) \\ = 7 \times 7 \\ = 49$

Answer 2

Answer:

(2-w)(2-W?)(2-w") (2

= (2 – w)(2- w)(2 - w x w")(2

= (2-w)(2-w)(2-wx 1)(2-w" x 1)

= (2 w)(2 w*)(2 w)(2

= (4 - 2w -2+w*)(4 - 2

= (4 - 2(w +w?) + 1)(4 2(w = (4 – 2(-1) + 1)(4- 2(-1

= (4 +2+1)(4 + 2 +1

= 7 x 7

= 49

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