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Express the following expression $$E=(x^3-y^3)(y^3-z^3)(z^3-x^3)$$ in terms of $a, b$ where $a,b \in \mathbb R$ and $$a=x^2y+y^2z+z^2x$$ $$b=xy^2+yz^2+zx^2$$

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    $\begingroup$ Have you tried multiplying it out? $\endgroup$ – Ross Millikan Feb 11 at 20:17
  • $\begingroup$ Yes, that’s right, I’ve wrote it wrong, my bad. $\endgroup$ – benj2k1 Feb 11 at 20:19
  • $\begingroup$ But how it could be solved, tho? $\endgroup$ – benj2k1 Feb 11 at 20:19
  • $\begingroup$ @RossMillikan already done that, it s just a mess a lot of powers and variable that doesn’t make sense in the end, it’s just a dead end. $\endgroup$ – benj2k1 Feb 11 at 20:21
  • $\begingroup$ Oh, come on people, this is really hard one. A specialy for someone new in this field. $\endgroup$ – Aqua Feb 11 at 20:32
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Firstly, we use $x^3-y^3=(x-y)(x^2+xy+y^2)$ for $(x,y)=(a,b),(b,c),(a,c)$ and we obtain $$E=(x-y)(y-z)(z-x)(x^2+xy+y^2)(y^2+yz+z^2)(x^2+xz+z^2)$$ $$E=(x^2z+y^2x+z^2y-x^2y-y^2z-z^2x)(x^2+xy+y^2)(y^2+yz+z^2)(x^2+xz+z^2)$$ $$E=(b-a)(x^2+xy+y^2)(y^2+yz+z^2)(x^2+xz+z^2)=(b-a)E^{'}$$ $$\text{Let's say }c=\frac{a}{xyz}=\frac{x}{z} + \frac {y}{x} + \frac {z}{y} \text{and }d=\frac{b}{xyz}=\frac{x}{y} + \frac {y}{z} + \frac {z}{x}$$ $$\frac{E^{'}}{(xyz)^2}=\prod_{cyc}{(x/yz+1/z+y/xz)}=(\sum_{cyc} \frac{x^2}{y^2} + 2 * \sum_{cyc} \frac{x}{z}) + ( \sum_{cyc} \frac{x^2}{z^2} + 2* \sum_{cyc} \frac{x}{y} ) + (\sum {xy/z^2} +3 ) $$ $$\text{Therefore, }\frac{E^{'}}{(xyz)^2}=c^2 + d^2 + cd=\frac{a^2+b^2+ab}{(xyz)^2}$$ $$\text{Now we have obtained }E^{'}=a^2+b^2+ab\text{, and we know, from earlier, }E=(b-a)E^{'}\text{, therefore}$$ $$E=(b-a)(a^2+b^2+ab) $$ $$E=b^3-a^3 $$

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  • $\begingroup$ That is, $E=b^3-a^3$ $\endgroup$ – saulspatz Feb 11 at 20:56
  • $\begingroup$ Indeed, thank you, I have added that now (at your suggestion). $\endgroup$ – Parallelism Alert Feb 11 at 20:57
  • $\begingroup$ I multiplied out $E$ and guessed the answer. I was trying to work out a way to prove it without doing all the multiplications, or resorting to a CAS, when you posted your answer. I may continue trying, though. $\endgroup$ – saulspatz Feb 11 at 21:00
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    $\begingroup$ Thank’s a lot, I thought through this way but I believed it will be a dead end, and I drop it. Thank you $\endgroup$ – benj2k1 Feb 11 at 21:02
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Start: $$ a-b = xy(x-y)+yz (y-z)+zx(z-x) $$ $$= xy(x-y)+yz (y-z)+zx(z-\color{red}y)+zx(\color{red}y-x) $$ $$= (xy-zx)(x-y)+(yz -zx)(y-z)$$

$$ =x(y-z)(x-y)+z(y-x)(y-z) $$ $$ = (x-y)(y-z)(x-z)$$

So $$ E = (a-b)\underbrace{(x^2+xy+y^2)(y^2+yz+z^2)(z^2+zx+x^2)}_{A}$$

Now you have to figer out $A$. (I bet it is $3ab$.)

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  • $\begingroup$ bro, it looks suspicious idk what to say about that A being 3ab $\endgroup$ – benj2k1 Feb 11 at 20:34
  • $\begingroup$ Did you calculate $A$. What about $3ab$? $\endgroup$ – Aqua Feb 11 at 20:35
  • $\begingroup$ I haven’t calculated A, cause it s a lot to process in and I think the answer to this question it s more elegant that all of those variables, idk I already try your version. That s why I m saying. $\endgroup$ – benj2k1 Feb 11 at 20:39
  • $\begingroup$ Please do cacluate $A$ and $3ab$ in OP othervise I will also vote for close. $\endgroup$ – Aqua Feb 11 at 20:40
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    $\begingroup$ @greedoid It is not 3ab. I've checked. $\endgroup$ – Don Thousand Feb 11 at 20:41

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