2011/048) show that (a+bw+cw^2) ^3 + (a + bw^2_+cw)^3 = (2a-b-c)(2b-a-c)(2c-a-b) when w is cube root of unity

let x = (a+bw+cw^2) ...1
and y = (a + bw^2+cw) ...2

we need to factor x^3 + y^3

x^3+y^3 = (x+y) ((x+y)^2 - 3 xy) ...3

add (2) and (1) to get

x+y = (2a+b(w+w^2)+ c(w+w^2))
= (2a -b - c) ... 4

now xy = (a+bw+cw^2)(a + bw^2+cw)
= (a^2 + b^2 + c^2 + ab(w+w^2) + bc(w+w^2) + ca(w+w^2))
= (a^2+b^2 + c^2 - ab - bc - ca)

so (x+y)^2 - 3xy
= (2a-b -c)^2 - 3(a^2+b^2 + c^2 - ab - bc - ca)
= (4a^2 + b^2 + c^2 - 4ab + 2bc - 4ac)- 3(a^2+b^2 + c^2 - ab - bc - ca)
= a^2 -2b^2 - 2c ^2 - ab - ac + 5 bc
= a^2 - a(b+c) - (2b^2 + 2c^2 - 5bc)
= a^2 -(a (2b-c) + (2c-b)) - (2b-c)(b-2c))
= a^2 -(a (2b-c) + (2c-b)) + (2b-c)(2c-b))
= (a - 2b + c )(a-2c + b)= (2b-a -c)(2 c-a -b )... 5


from 3 4 and 5 we get the result