Question

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

Answer 1

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.

`xy(x^2-y^2) +yz(y^2-z^2) +zx(z^2-x^2)`

`=xy(x+y)(x-y) -yz(y+z)(y-z) +zx(z+x)(z-x)`

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.

Answer 2

`(x - y) (x - z) (y - z) (x + y + z)`

Explanation:

`f=x y (x^2 - y^2) + y z (y^2 - z^2) + z x (z^2 - x^2)`

Calling `y = lambda x` and `x = mu x` and substituting

`f=(lambda - lambda^3 - mu + lambda^3 mu + mu^3 - lambda mu^3)x^4`

Now examining the polynomial equation

`g(lambda,mu)=lambda - lambda^3 - mu + lambda^3 mu + mu^3 - lambda mu^3=0`

we can verify that

`g(1,1)=0` and
`g(lambda,lambda)=g(mu,mu)=0` so

`g(lambda,mu) = (lambda-1)(mu-1)(lambda-mu)(a lambda+bmu+c)`

Now expanding and comparing coefficients

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

`f=(lambda-1)(mu-1)(lambda-mu)( lambda+mu+1)x^4` or

`f = (lambda x - x)(mu x - x)(lambda x-mu x)(lambda x+mu x + x)` or

`f=(y-x)(z-x)(y-z)(y+z+x)`

Finally

`x y (x^2 - y^2) + y z (y^2 - z^2) + z x (z^2 - x^2)=(x - y) (x - z) (y - z) (x + y + z)`

Answer 3

`x y (x^2 - y^2) + y z (y^2 - z^2) + z x (z^2 - x^2)`

`=x y (x^2 - y^2) + y^3z - yz^3 + z^3x - zx^3`

`=x y (x^2 - y^2) -zx^3+ y^3z + z^3x- yz^3`

`=x y (x^2 - y^2) -z(x^3- y^3) + z^3(x- y)`

`=x y (x+y)(x - y) -z(x- y)(x^2+xy+y^2) + z^3(x- y)`

`=(x-y)[x y (x+y) -z(x^2+xy+y^2) + z^3]`

`=(x-y)[x^2y +xy^2 -zx^2-xyz-y^2z + z^3]`

`=(x-y)[x^2y -zx^2+xy^2 -xyz-y^2z + z^3]`

`=(x-y)[x^2(y -z)+xy(y -z)-z(y^2 -z^2)]`

`=(x-y)(y-z)[x^2+xy-zy -z^2]`

`=(x-y)(y-z)[x^2 -z^2+xy-zy ]`

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

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