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I have to do the following problem:

If $ax+cy+bz = X, cx+by+az=Y, bx+ay+cz=Z$, then show that

$(a^2+b^2+c^2-bc-ca-ab)(x^2+y^2+z^2-yz-zx-xy) = X^2+Y^2+Z^2-YZ-XZ-XY$.

It's easy enough through normal Algebra if I pair $X^2-XY = X(X-Y)$ etc., and collect the coefficients on both sides, but the chapter is of complex numbers and more specifically, cube roots of unity. How do I solve this using that idea?

The best I've come up with is to replace $-1$ with $\omega+\omega^2$, but I later seem to apply the reverse and thus, render the use of $\omega$ pointless. Some guidance, please!

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ankush981
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1 Answers1

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HINT:

$$\sum_{\text{cyc}}(X^2-YZ)=X^2+Y^2+Z^2+(XY+YZ+ZX)(w^2+w)$$

$$=(X+Yw+Zw^2)(X+Yw^2+Zw)$$ where $w$ is a complex cube root of $1$

$$X+Yw+Zw^2=ax+cy+bz +w(cx+by+az)+w^2(bx+ay+cz)$$

$$=a(x+w^2y+wz)+bw^2(x+y^2w+wz)+cw(x+y^2w+wz)=(x+y^2w+wz)(a+bw^2+cw)$$

$$\text{Similarly, for } X+Yw^2+Zw$$

Alternatively, without using Complex Numbers,

$$X^2-YZ=(ax+cy+bz)^2-(cx+by+az)(bx+ay+cz)$$

$$=(a^2-bc)x^2+(c^2-ab)y^2+(b^2-ca)z^2+(2ca-b^2-ca)xy+(2ab-c^2-bc)zx+(2bc-ab-a^2)yz$$

$$\implies\sum_{\text{cyc}}(X^2-YZ)=(x^2+y^2+z^2-xy-yz-zx)(a^2+b^2+c^2-ab-bc-ca)$$

lab bhattacharjee
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  • @dotslash, which book is it? Also, you have made wrong pairing – lab bhattacharjee Aug 10 '13 at 15:13
  • Higher Algebra by Hall & Knight. It's the last problem in the chapter "Surds and Imaginary Quantities". – ankush981 Aug 10 '13 at 15:15
  • Wow, thanks! But how did you arrive the second step? Intuition? It was a magnificent stroke of reduction! :) – ankush981 Aug 10 '13 at 15:24
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    @dotslash, my pleasure. Frankly speaking once, I proved that $a+bw+cw^2$ is a factor of $\sum a^3-3abc$ as followed: $$a+bw+cw^2=0\iff -a=bw+cw^2$$ Cubing we get, $$-a^3=b^3+c^3+3bw\cdot cw^2(bw+cw^2)=b^3+c^3+3bc(-a)$$ – lab bhattacharjee Aug 10 '13 at 15:34
  • @dotslash, try putting $w=1$ in my last comment. Also observe, $X+Y+Z=\cdots=(a+b+c)(x+y+z)$ so $\sum_{\text{cyc}}X^3-3XYZ=(\sum_{\text{cyc}}x^3-3xyz)(\sum_{\text{cyc}}a^3-3abc)$ – lab bhattacharjee Aug 10 '13 at 15:56
  • @labbhattacharjee Your last example is a bit hard to understand, but I'll try wrapping my head around it in leisure. Thanks a lot for the inputs. :) – ankush981 Aug 10 '13 at 16:01
  • @dotslash, can you factorize $a^3+b^3+c^3-3abc?$ – lab bhattacharjee Aug 10 '13 at 16:14
  • @labbhattacharjee Given this background, I'm tempted to say $\Pi (a+b \omega + c \omega^2)$ (multiplying these three terms should tell whether that's correct or not), but I don't think my skills are developed enough to arrived at it independently. In other words, I can't factorize it without guesswork! I spent some time doing similar multiplications last night, just to familiarize myself with the importance of $\omega + \omega^2$ in factorization, but I'm not sure how good I am. :D – ankush981 Aug 11 '13 at 07:19