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Grade: 12

                        

Prove that (1-w)(1-w^2)(1-w^4)(1-w^8)=9 in algebra method.

Prove that (1-w)(1-w^2)(1-w^4)(1-w^8)=9 in algebra method.

3 years ago

Answers : (4)

Arun
25768 Points
							
 
We have 1 + w + w2 = 0 & w3 = 1
(1 – w)(1 – w2)(1 – w4)(1 – w5) = (1 – w)(1 – w2)(1 – w3.w)(1 – w3.w2)
= (1 – w)(1 – w2)(1 – w)(1 – w2)  (Since w3 = 1)
= (1 – w)2(1 – w2)2
=(1 – 2w + w2)(1 – 2w2 + w4) = (1 – 2w + w2) (1 – 2w2 – w3.w) = (1 – 2w + w2) (1 – 2w2 – w) 
= ( – w – 2w)( – w2 – 2w2)  (since 1 + w + w2 = 0)
= ( – 3w) ( – 3w2) = 9w3 = 9
3 years ago
Rishi Sharma
askIITians Faculty
629 Points
							Dear Student,
Please find below the solution to your problem.

We have 1 + w + w2= 0 & w3= 1
(1 – w)(1 – w2)(1 – w4)(1 – w5) = (1 – w)(1 – w2)(1 – w3.w)(1 – w3.w2)
= (1 – w)(1 – w2)(1 – w)(1 – w2) (Since w3= 1)
= (1 – w)2(1 – w2)2
=(1 – 2w + w2)(1 – 2w2+ w4) = (1 – 2w + w2) (1 – 2w2– w3.w) =(1 – 2w + w2) (1 – 2w2– w)
= ( – w – 2w)( – w2– 2w2) (since1 + w + w2= 0)
= ( – 3w) ( – 3w2) = 9w3= 9

Thanks and Regards
5 months ago
Shanmukha Rao
27 Points
							
We have 1 + w + w^2= 0 & w^3= 1=(1 – w)(1 – w^2)(1 – w^4)(1 – w^5) = (1 – w)(1 – w^2)(1 – w^3.w)(1 – w^3.w^2)= (1 – w)(1 – w^2)(1 – w)(1 – w^2)                         (Since w^3= 1)= (1 – w)^2(1 – w^2)^2=(1 – 2w + w2)(1 – 2w2+ w4) = (1 – 2w + w^2) (1 – 2w^2– w^3.w) =(1 – 2w + w^2) (1 – 2w^2– w)= ( – w – 2w)( – w^2– 2w^2)                 (since1 + w + w^2= 0)= ( – 3w) ( – 3w^2) = 9w^3= 9
one month ago
Shanmukha Rao
27 Points
							
We have: 1 + w + w^2= 0 & w^3= 1=(1 – w)(1 – w^2)(1 – w^4)(1 – w^5) = (1 – w)(1 – w^2)(1 – w^3.w)(1 – w^3.w^2)= (1 – w)(1 – w^2)(1 – w)(1 – w^2)                         (Since w^3= 1)= (1 – w)^2(1 – w^2)^2=(1 – 2w + w2)(1 – 2w2+ w4) = (1 – 2w + w^2) (1 – 2w^2– w^3.w) =(1 – 2w + w^2) (1 – 2w^2– w)= ( – w – 2w)( – w^2– 2w^2)                 (since1 + w + w^2= 0)= ( – 3w) ( – 3w^2) = 9w^3= 9
one month ago
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