Algebra> Prove that (1-w)(1-w^2)(1-w^4)(1-w^8)=9 i...

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

Srichandan , 8 Years ago

Grade 12

5 Answers

Arun

Last Activity: 8 Years ago

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

Rishi Sharma

Last Activity: 5 Years ago

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

Shanmukha Rao

Last Activity: 5 Years ago

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

Shanmukha Rao

Last Activity: 5 Years ago

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