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?

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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?

godsrvnt0706 · Accepted Answer

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 t3−1:  t3−1=(t−1)(t−w)(t−w2)  And then substituting t=x−y on both sides.  Now, using 1+w+w2=0 (sum of roots in t3−1=0 by Vietta) repeatdly, observe that:  1) X+Y=2a−b−c=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 · Accepted Answer

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)2−3(ab+bc+ca)  =−3(ab+bc+ca) since a+b+c=0  =−3(a(b+c)+bc)  =−3a2−3bc since b+c=−a  U3+V3=(U+V)3−3UV(U+V)  =(3a)3−3(−3a2−3bc)⋅3a  =27a3−27a3+27abc  =27abc

If a+b+c=0, and w is a complex root of cube

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