# How do you factorize xy(x^2-y^2)+yz(y^2-z^2)+zx(z^2-x^2) ?

Jun 17, 2017

Despite the fact that the same variables appear in every term, this expression does not factorize into one term.

There is no common factor, and no common bracket.

EAch term can be factored by differemce of squares.

$x y \left({x}^{2} - {y}^{2}\right) + y z \left({y}^{2} - {z}^{2}\right) + z x \left({z}^{2} - {x}^{2}\right)$

$= x y \left(x + y\right) \left(x - y\right) - y z \left(y + z\right) \left(y - z\right) + z x \left(z + x\right) \left(z - x\right)$

The only other option would be to multiply out the brackets and try a different grouping, but I don not think it will produce a better result than this.

Jun 17, 2017

$\left(x - y\right) \left(x - z\right) \left(y - z\right) \left(x + y + z\right)$

#### Explanation:

$f = x y \left({x}^{2} - {y}^{2}\right) + y z \left({y}^{2} - {z}^{2}\right) + z x \left({z}^{2} - {x}^{2}\right)$

Calling $y = \lambda x$ and $x = \mu x$ and substituting

$f = \left(\lambda - {\lambda}^{3} - \mu + {\lambda}^{3} \mu + {\mu}^{3} - \lambda {\mu}^{3}\right) {x}^{4}$

Now examining the polynomial equation

$g \left(\lambda , \mu\right) = \lambda - {\lambda}^{3} - \mu + {\lambda}^{3} \mu + {\mu}^{3} - \lambda {\mu}^{3} = 0$

we can verify that

$g \left(1 , 1\right) = 0$ and
$g \left(\lambda , \lambda\right) = g \left(\mu , \mu\right) = 0$ so

$g \left(\lambda , \mu\right) = \left(\lambda - 1\right) \left(\mu - 1\right) \left(\lambda - \mu\right) \left(a \lambda + b \mu + c\right)$

Now expanding and comparing coefficients

$a = 1 , b = 1 , c = 1$ so finally

$f = \left(\lambda - 1\right) \left(\mu - 1\right) \left(\lambda - \mu\right) \left(\lambda + \mu + 1\right) {x}^{4}$ or

$f = \left(\lambda x - x\right) \left(\mu x - x\right) \left(\lambda x - \mu x\right) \left(\lambda x + \mu x + x\right)$ or

$f = \left(y - x\right) \left(z - x\right) \left(y - z\right) \left(y + z + x\right)$

Finally

$x y \left({x}^{2} - {y}^{2}\right) + y z \left({y}^{2} - {z}^{2}\right) + z x \left({z}^{2} - {x}^{2}\right) = \left(x - y\right) \left(x - z\right) \left(y - z\right) \left(x + y + z\right)$

Jul 29, 2018

$x y \left({x}^{2} - {y}^{2}\right) + y z \left({y}^{2} - {z}^{2}\right) + z x \left({z}^{2} - {x}^{2}\right)$

$= x y \left({x}^{2} - {y}^{2}\right) + {y}^{3} z - y {z}^{3} + {z}^{3} x - z {x}^{3}$

$= x y \left({x}^{2} - {y}^{2}\right) - z {x}^{3} + {y}^{3} z + {z}^{3} x - y {z}^{3}$

$= x y \left({x}^{2} - {y}^{2}\right) - z \left({x}^{3} - {y}^{3}\right) + {z}^{3} \left(x - y\right)$

$= x y \left(x + y\right) \left(x - y\right) - z \left(x - y\right) \left({x}^{2} + x y + {y}^{2}\right) + {z}^{3} \left(x - y\right)$

$= \left(x - y\right) \left[x y \left(x + y\right) - z \left({x}^{2} + x y + {y}^{2}\right) + {z}^{3}\right]$

$= \left(x - y\right) \left[{x}^{2} y + x {y}^{2} - z {x}^{2} - x y z - {y}^{2} z + {z}^{3}\right]$

$= \left(x - y\right) \left[{x}^{2} y - z {x}^{2} + x {y}^{2} - x y z - {y}^{2} z + {z}^{3}\right]$

$= \left(x - y\right) \left[{x}^{2} \left(y - z\right) + x y \left(y - z\right) - z \left({y}^{2} - {z}^{2}\right)\right]$

$= \left(x - y\right) \left(y - z\right) \left[{x}^{2} + x y - z y - {z}^{2}\right]$

$= \left(x - y\right) \left(y - z\right) \left[{x}^{2} - {z}^{2} + x y - z y\right]$

$= \left(x - y\right) \left(y - z\right) \left[\left(x - z\right) \left(x + z\right) + y \left(x - z\right)\right]$

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