Complex Roots Of Unity Q. (1 Viewer)

[kevda1st]

Prove that if: a = x+ y b=xw+yw² c=xw²+yw^4 ,
Than abc = x³+y³ and a²+b²+c²=6xy


and part ii) of this Question....
Thanks in advance

 

[Trebla]

and part ii) of this Question....
Thanks in advance

(z + √3)⁶ + 64 = 0
Divide by 64
[(z + √3)/2]⁶ + 1 = 0
Let x = (z + √3)/2
=> x⁶ + 1 = 0
The solutions for x are the solutions found in part (i). Therefore the solutions of z are found by the solution set of x and then substituting into z = 2x - √3

 

[Timothy.Siu]

Prove that if: a = x+ y b=xw+yw² c=xw²+yw^4 ,
Than abc = x³+y³ and a²+b²+c²=6xy

i guess that w is a root of unity of w^3=1 then, 1+w+w^2=0
abc=(x+y)(xw+yw^2)(xw^2+yw^4)
=(x+y)(x^2w^3+xyw^5+xyw^4+y^2w^6)
=(x+y)(x^2+xy(w^2+w)+y^2)
=(x+y)(x^2-xy+y^2)=x^3+y^3

a^2+b^2+c^2=(a+b+c)^2-2(ab+bc+ca)