If x=a+b, y=aw+bw^2,z=aw^2+bw, prove that x^3+y^3+z^3=3(a^3+b^3)
Answers
Answer:
download snapsolve app or doubtnut app from playstore to know your answer
Answer:
A knowledge of enough algebra, at least to know why these three things are true.
a+b+c=0⟹a3+b3+c3=3abc
“Any equation of nth degree has at most n roots, therefore unity or for that matter any number has 3 cube roots”
And that
a3±b3=(a±b)(a2∓ab+b2)
The cube roots of unity are solutions to the equation
x3=1
⟹x3−1=0
⟹(x−1)(x2+x+1)=0
which means a cube root of unity either follows
x−1=0or
x2+x+1=0
HINT: This second property shall help you solve this; do not look at what follows before trying it yourself.
We have this information
x=a+b,y=aω+bω2,z=aω2+bω
Since,
x+y+z=a+aω+aω2+b+bω+bω2
⟹x+y+z=(a+b)(1+ω+ω2)
⟹x+y+z=0
Now, the proof is easy!
x3+y3+z3=3xyz
=3(a+b)(aω+bω2)(aω2+bω)
=3(a+b)(a2+abω2+abω+b2)
=3(a+b)(a2−ab+b2)
Therefore
x3+y3+z3=3(a3+b3)