MHB Algebra Challenge: Express $(x^3-y^3)(y^3-z^3)(z^3-x^3)$ in Terms of a & b

[anemone]

Let $a,\,b,\,x,\,y,\,z$ be real numbers such that $x^2y+y^2z+z^2x=a$ and $xy^2+yz^2+zx^2=b$.

Express $(x^3-y^3)(y^3-z^3)(z^3-x^3)$ in terms of $a$ and $b$.

 

[Albert1]

Let $a,\,b,\,x,\,y,\,z$ be real numbers such that $x^2y+y^2z+z^2x=a$ and $xy^2+yz^2+zx^2=b$.

Express $(x^3+y^3)(y^3-z^3)(z^3-x^3)$ in terms of $a$ and $b$.

first term:$(x^3+y^3)--? $ or $(x^3-y^3)-- ? $

 

[anemone]

first term:$(x^3+y^3)--? $ or $(x^3-y^3)-- ? $

Ops...You're right, the plus should be a minus sign...sorry:(, I will edit my original post now.

 

[anemone]

Let $a,\,b,\,x,\,y,\,z$ be real numbers such that $x^2y+y^2z+z^2x=a$ and $xy^2+yz^2+zx^2=b$.

Express $(x^3-y^3)(y^3-z^3)(z^3-x^3)$ in terms of $a$ and $b$.

Hint:

Spoiler

Cube root of unity is used to simplify the problem.

 

[kaliprasad]

not solved since long
here I close it(it took me lot of time may be 20hrs of thought)

Spoiler

$(x^3-y^3)(y^3-z^3)(z^3-x^3)$
= $(x^3-y^3)(y^3z^3-y^3x^3-z^6+z^3x^3)$
=$x^3y^3z^3-y^3x^6-z^6x^3+z^3x^6-y^6z^3+y^6x^3+y^3z^6-x^3z^3x^3$
= $-y^3x^6-z^6x^3+z^3x^6-y^6z^3+y^6x^3+y^3z^6$
= $(z^3x^6+y^3 z^6 + x^3y^6) - (y^3x^6 + x^3z^6 + y^6z^3)$
hence
$(x^3-y^3)(y^3-z^3)^(z^3-x^3)= ((xy^2)^3 + (yz^2)^3+(zx^2)^3) - ((x^2y)^3 +(y^2z)^3 + (z^2x)^3)\cdots(1)$
we know
$a^3+b^3 + c^3 = (a+b+c)^3 - 3(a+b)(b+c)(c+a)$
so $((xy^2)^3 + (yz^2)^3+(zx^2)^3) = (xy^2+yz^2+zx^2)^3- 3(xy^2+yz^2)(yz^2+zx^2)(zx^2+xy^2)$

= $(xy^2+yz^2+zx^2) - 3y(xy+z^2)z(x^2+yz)x(xz+y^2)$

= $b^2-3xyz(x^2+yz)(y^2+xz)(z^2+xy)\cdots(2)$

similarly

$((x^2y)^3 + (y^2z)^3+(z^2x)^3) =a^2-3xyz(x^2+yz)(y^2+xz)(z^2+xy)\cdots(3)$

from (1), (2) and (3) we get

$(x^3-y^3)(y^3-z^3)^(z^3-x^3)=b^3 - a^3 $

I would like to see the solution with the hint.

 

[Opalg]

I would like to see the solution with the hint.

Spoiler

Let $\omega$ be a complex cube root of unity, as in anemone's hint. Then $$(y-z)(z-x)(x-y) = b-a,$$ $$(y-\omega z)(z-\omega x)(x-\omega y) = \omega^2 b-\omega a = \omega^2(b - \omega^2 a),$$ $$(y-\omega^2 z)(z-\omega^2 x)(x-\omega^2 y) = \omega b-\omega^2 a = \omega(b - \omega a).$$

Therefore $$\begin{aligned}(y^3 - z^3)(z^3 - x^3)(x^3 - y^3) &= (y-z)(y-\omega z)(y-\omega^2z)\, (z-x)(z-\omega x)(z-\omega^2x)\, (x-y)(x-\omega y)(x-\omega^2y) \\ &= (y-z)(z-x)(x-y)\, (y-\omega z)(z-\omega x)(x-\omega y)\, (y-\omega^2 z)(z-\omega^2 x)(x-\omega^2 y) \\ &= (b-a)\omega^2 (b-\omega a)\omega(b-\omega a) \\ &= \omega^3(b-a)(b-\omega a)(b-\omega^2 a) = b^3 - a^3. \end{aligned}$$