Grade 11Algebra
If 1, w, w² are cube roots of unity, then prove that (3+3w+5w²)6 - (2+6w+2w²)3 = 0.
If 1, w, w² are cube roots of unity, then prove that (3+3w+5w²)6 - (2+6w+2w²)3 = 0.
1 Answer
Vikas TU
Dear student
w is cube rot of unity ,
So w^2 + w+1 = 0
(3+3w+5w^2)^6 - (2+6w+2w^2)^3
= (3+3w+3w^2 +2w^2)^6 - (2+ 2w + 4w + 2w^2)^3
= 64w^2 - 64w^2 = 0
Hope this helps Dear student
w is cube rot of unity ,
So w^2 + w+1 = 0
(3+3w+5w^2)^6 - (2+6w+2w^2)^3
= (3+3w+3w^2 +2w^2)^6 - (2+ 2w + 4w + 2w^2)^3
= 64w^2 - 64w^2 = 0
Hope this helps
w is cube rot of unity ,
So w^2 + w+1 = 0
(3+3w+5w^2)^6 - (2+6w+2w^2)^3
= (3+3w+3w^2 +2w^2)^6 - (2+ 2w + 4w + 2w^2)^3
= 64w^2 - 64w^2 = 0
Hope this helps Dear student
w is cube rot of unity ,
So w^2 + w+1 = 0
(3+3w+5w^2)^6 - (2+6w+2w^2)^3
= (3+3w+3w^2 +2w^2)^6 - (2+ 2w + 4w + 2w^2)^3
= 64w^2 - 64w^2 = 0
Hope this helps
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