If a+b+c=0, and w is a complex root of cube roots of unity, then can...

Patricia Crane

Patricia Crane

Answered

2022-01-17

If a+b+c=0, and w is a complex root of cube roots of unity, then can you show that
(a+bw+cw2)3+(a+bw2+cw)3=27abc?

Answer & Explanation

godsrvnt0706

godsrvnt0706

Expert

2022-01-18Added 31 answers

Let X=(a+bw+cw2) and Y=(a+bw2+cw).
The key idea is to note that
X3+Y3=(X+Y)(X+wY)(X+w2Y) for all X,Y
This, in turn, follows from using the cube roots of unit (1,w,w2) to factor t31:
t31=(t1)(tw)(tw2)
And then substituting t=xy on both sides.
Now, using 1+w+w2=0 (sum of roots in t31=0 by Vietta) repeatdly, observe that:
1) X+Y=2abc=3a
2) X+wY=a+bw+cw2+w(a+bw2+cw)=a(1+w)+b(w+1)+2cw2=3cw2
3) X+w2Y=a+bw+cw2+w2(a+bw2+cw)=a(1+w2)+2bw+c(w2+1)=3bw
Multiplying out (i), (ii) and (iii), we get:
X3+Y3=(X+Y)(X+wY)(X+w2Y)=27abcw3=27abc
ol3i4c5s4hr

ol3i4c5s4hr

Expert

2022-01-19Added 48 answers

Let U=a+bω+cw2 and V=a+bω2+cω
UV=a2+(ab+ac)ω+(ab+bc+ca)ω2+b2ω3+c2ω3+bcω4
=a2+b2+c2+(ab+bc+ca)(ω+ω2) since ω4=ω
=a2+b2+c2(ab+bc+ca)
=(a+b+c)23(ab+bc+ca)
=3(ab+bc+ca) since a+b+c=0
=3(a(b+c)+bc)
=3a23bc since b+c=a
U3+V3=(U+V)33UV(U+V)
=(3a)33(3a23bc)3a
=27a327a3+27abc
=27abc

Do you have a similar question?

Recalculate according to your conditions!

Ask your question.
Get your answer.

Let our experts help you. Answer in as fast as 15 minutes.

Didn't find what you were looking for?