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If w is a non-real cube root of 1, then(2+3w+2w^2)^5+(2+3w^2+2w)^5 = ?

Sanjali , 6 Years ago
Grade 12
anser 2 Answers
Khimraj

Last Activity: 6 Years ago

5+2w)]2+ (2+2w+[w5)]2 w +(2+2w+2wLet I = (2+3w+2w2)5+(2+3w2+2w)5= [

SInce 1+w+w2 = 0
SO 
I = w5 + w10
Since w3 = 1
SO 
I = w2 + w 
SInce 1+w+w2 = 0
So
I = -1.
Hope it clears. If you like answer then please approve it.

 

Khimraj

Last Activity: 6 Years ago

First step is misprinted So i am writting it again
Let I = (2 + 3w + 2w2)5 + (2 + 3w2 + 2w)5
So I = [ w + (2 + 2w + 2w2)]5 + [ w2 + (2+ 2w2 + 2w)]5

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