Patricia Crane

2022-01-17

If $a+b+c=0$, and w is a complex root of cube roots of unity, then can you show that
${\left(a+bw+c{w}^{2}\right)}^{3}+{\left(a+b{w}^{2}+cw\right)}^{3}=27abc$?

godsrvnt0706

Expert

Let .
The key idea is to note that
${X}^{3}+{Y}^{3}=\left(X+Y\right)\left(X+wY\right)\left(X+{w}^{2}Y\right)$ for all X,Y
This, in turn, follows from using the cube roots of unit $\left(1,w,{w}^{2}\right)$ to factor ${t}^{3}-1$:
${t}^{3}-1=\left(t-1\right)\left(t-w\right)\left(t-{w}^{2}\right)$
And then substituting $t=\frac{x}{-y}$ on both sides.
Now, using $1+w+{w}^{2}=0$ (sum of roots in ${t}^{3}-1=0$ by Vietta) repeatdly, observe that:
1) $X+Y=2a-b-c=3a$
2) $X+wY=a+bw+c{w}^{2}+w\left(a+b{w}^{2}+cw\right)=a\left(1+w\right)+b\left(w+1\right)+2c{w}^{2}=3c{w}^{2}$
3) $X+{w}^{2}Y=a+bw+c{w}^{2}+{w}^{2}\left(a+b{w}^{2}+cw\right)=a\left(1+{w}^{2}\right)+2bw+c\left({w}^{2}+1\right)=3bw$
Multiplying out (i), (ii) and (iii), we get:
${X}^{3}+{Y}^{3}=\left(X+Y\right)\left(X+wY\right)\left(X+{w}^{2}Y\right)=27abc{w}^{3}=27abc$

ol3i4c5s4hr

Expert

Let
$UV={a}^{2}+\left(ab+ac\right)\omega +\left(ab+bc+ca\right){\omega }^{2}+{b}^{2}{\omega }^{3}+{c}^{2}{\omega }^{3}+bc{\omega }^{4}$

$={a}^{2}+{b}^{2}+{c}^{2}-\left(ab+bc+ca\right)$
$={\left(a+b+c\right)}^{2}-3\left(ab+bc+ca\right)$

$=-3\left(a\left(b+c\right)+bc\right)$

${U}^{3}+{V}^{3}={\left(U+V\right)}^{3}-3UV\left(U+V\right)$
$={\left(3a\right)}^{3}-3\left(-3{a}^{2}-3bc\right)\cdot 3a$
$=27{a}^{3}-27{a}^{3}+27abc$
$=27abc$

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