Question

if w is a complex cube root of unity such that x-a+b,y=a+aw² and z-aw²+be,a,b€R
then prove that x+y+z=0​

Answer 1

Answer:

I am assuming w=ω=1–√3

In that case we are expected to prove

if x=a+b,y=aω+bω2,z=aω2+bω

then x3+b3+z3=3(a3+b3)

where we know

1+ω+ω2=0

a3+b3=3(a+b)(a2−ab+b2)

if a+b+c=0 then a3+b3+c3=3abc

now,

x+y+z=a+b+(aω+bω2)+aω2+bω

⇒x+y+z=(a+b)(1+ω+ω2)=0

and we know if x+y+z=0 , then x3+y3+z3=3xyz

⇒x3+y3+z3= 3(a+b)(aω+bω2)(aω2+bω)

⇒x3+y3+z3=3ω2(a+b)(a+bω)(aω+b)

⇒x3+y3+z3= 3ω2(a+b)(a2ω+ab(1+ω2)+b2ω)

⇒x3+y3+z3=3ω2(a+b)(a2ω−abω+b2)

⇒x3+y3+z3=3ω3(a+b)(a2−ab+b2)

⇒x3+y3+z3=3(a3+b3)