If 1 2 , w, w are the cube roots of unity, then prove the following : (1+w) (1+w^2 ) (1+w^7 ) (1 +w^*8)=1

Question

Ritesh Khatri · Answer

we know that w 3 = 1 and 1 + w + w 2 = 0 therefore w 7 = w and w 8 = w 2 given expression becomes (1+w) 2 (1+w 2 ) 2 =[(1+w) (1+w 2 )] 2 =(1 + w 3 + w + w 2 ) 2 =(1 + 0 ) 2 =1

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If 1 2 , w, w are the cube roots of unity, then prove the following : (1+w) (1+w^2 ) (1+w^7 ) (1 +w^*8)=1

If 1 2 , w, w are the cube roots of unity, then prove the following :
(1+w) (1+w^2 ) (1+w^7 ) (1 +w^*8)=1

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If 1 2 , w, w are the cube roots of unity, then prove the following : (1+w) (1+w^2 ) (1+w^7 ) (1 +w^*8)=1

If 1 2 , w, w are the cube roots of unity, then prove the following :(1+w) (1+w^2 ) (1+w^7 ) (1 +w^*8)=1

1 Answers

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If 1 2 , w, w are the cube roots of unity, then prove the following :
(1+w) (1+w^2 ) (1+w^7 ) (1 +w^*8)=1