If 1, w, w² are cube roots of unity, then prove that (3+3w+5w²)6 - (2+6w+2w²)3 = 0.

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