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

${(a+bw+c{w}^{2})}^{3}+{(a+b{w}^{2}+cw)}^{3}=27abc$ ?

Answer & Explanation

godsrvnt0706

Expert

2022-01-18Added 31 answers

Let $X=(a+bw+c{w}^{2})\text{}\text{and}\text{}Y=(a+b{w}^{2}+cw)$ .

The key idea is to note that

${X}^{3}+{Y}^{3}=(X+Y)(X+wY)(X+{w}^{2}Y)$ for all X,Y

This, in turn, follows from using the cube roots of unit$(1,w,{w}^{2})$ to factor ${t}^{3}-1$ :

${t}^{3}-1=(t-1)(t-w)(t-{w}^{2})$

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)$X+wY=a+bw+c{w}^{2}+w(a+b{w}^{2}+cw)=a(1+w)+b(w+1)+2c{w}^{2}=3c{w}^{2}$

3)$X+{w}^{2}Y=a+bw+c{w}^{2}+{w}^{2}(a+b{w}^{2}+cw)=a(1+{w}^{2})+2bw+c({w}^{2}+1)=3bw$

Multiplying out (i), (ii) and (iii), we get:

${X}^{3}+{Y}^{3}=(X+Y)(X+wY)(X+{w}^{2}Y)=27abc{w}^{3}=27abc$

The key idea is to note that

This, in turn, follows from using the cube roots of unit

And then substituting

Now, using

1)

2)

3)

Multiplying out (i), (ii) and (iii), we get:

ol3i4c5s4hr

Expert

2022-01-19Added 48 answers

Let $U=a+b\omega +c{w}^{2}\text{}\text{and}\text{}V=a+b{\omega}^{2}+c\omega$

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

$={a}^{2}+{b}^{2}+{c}^{2}+(ab+bc+ca)(\omega +{\omega}^{2})\text{}\text{since}\text{}{\omega}^{4}=\omega$

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

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

$=-3(ab+bc+ca)\text{}\text{since}\text{}a+b+c=0$

$=-3(a(b+c)+bc)$

$=-3{a}^{2}-3bc\text{}\text{since}\text{}b+c=-a$

${U}^{3}+{V}^{3}={(U+V)}^{3}-3UV(U+V)$

$={\left(3a\right)}^{3}-3(-3{a}^{2}-3bc)\cdot 3a$

$=27{a}^{3}-27{a}^{3}+27abc$

$=27abc$

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