# How to solve this algebraic manipulation problem?

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 $$a=x^2y+y^2z+z^2x$$ $$b=xy^2+yz^2+zx^2$$

• Have you tried multiplying it out? – Ross Millikan Feb 11 at 20:17
• Yes, that’s right, I’ve wrote it wrong, my bad. – benj2k1 Feb 11 at 20:19
• But how it could be solved, tho? – benj2k1 Feb 11 at 20:19
• @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. – benj2k1 Feb 11 at 20:21
• Oh, come on people, this is really hard one. A specialy for someone new in this field. – Aqua Feb 11 at 20:32

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 $$E=(x-y)(y-z)(z-x)(x^2+xy+y^2)(y^2+yz+z^2)(x^2+xz+z^2)$$ $$E=(x^2z+y^2x+z^2y-x^2y-y^2z-z^2x)(x^2+xy+y^2)(y^2+yz+z^2)(x^2+xz+z^2)$$ $$E=(b-a)(x^2+xy+y^2)(y^2+yz+z^2)(x^2+xz+z^2)=(b-a)E^{'}$$ $$\text{Let's say }c=\frac{a}{xyz}=\frac{x}{z} + \frac {y}{x} + \frac {z}{y} \text{and }d=\frac{b}{xyz}=\frac{x}{y} + \frac {y}{z} + \frac {z}{x}$$ $$\frac{E^{'}}{(xyz)^2}=\prod_{cyc}{(x/yz+1/z+y/xz)}=(\sum_{cyc} \frac{x^2}{y^2} + 2 * \sum_{cyc} \frac{x}{z}) + ( \sum_{cyc} \frac{x^2}{z^2} + 2* \sum_{cyc} \frac{x}{y} ) + (\sum {xy/z^2} +3 )$$ $$\text{Therefore, }\frac{E^{'}}{(xyz)^2}=c^2 + d^2 + cd=\frac{a^2+b^2+ab}{(xyz)^2}$$ $$\text{Now we have obtained }E^{'}=a^2+b^2+ab\text{, and we know, from earlier, }E=(b-a)E^{'}\text{, therefore}$$ $$E=(b-a)(a^2+b^2+ab)$$ $$E=b^3-a^3$$

• That is, $E=b^3-a^3$ – saulspatz Feb 11 at 20:56
• Indeed, thank you, I have added that now (at your suggestion). – Parallelism Alert Feb 11 at 20:57
• 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. – saulspatz Feb 11 at 21:00
• Thank’s a lot, I thought through this way but I believed it will be a dead end, and I drop it. Thank you – benj2k1 Feb 11 at 21:02

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 $$3ab$$.)

• bro, it looks suspicious idk what to say about that A being 3ab – benj2k1 Feb 11 at 20:34
• Did you calculate $A$. What about $3ab$? – Aqua Feb 11 at 20:35
• 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. – benj2k1 Feb 11 at 20:39
• Please do cacluate $A$ and $3ab$ in OP othervise I will also vote for close. – Aqua Feb 11 at 20:40
• @greedoid It is not 3ab. I've checked. – Don Thousand Feb 11 at 20:41