Question

If x = a + b, y = αa + βb and z=aβ + bα where α and β are complex cube-roots of unity, show that xyz = a³ + b³.

Answer 1

Answer:

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)

Answer 2

Answer:the best and easiest way

Step-by-step explanation:

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