How do you factorize xy(x2y2)+yz(y2z2)+zx(z2x2) ?

3 Answers
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.

xy(x2y2)+yz(y2z2)+zx(z2x2)

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

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

(xy)(xz)(yz)(x+y+z)

Explanation:

f=xy(x2y2)+yz(y2z2)+zx(z2x2)

Calling y=λx and x=μx and substituting

f=(λλ3μ+λ3μ+μ3λμ3)x4

Now examining the polynomial equation

g(λ,μ)=λλ3μ+λ3μ+μ3λμ3=0

we can verify that

g(1,1)=0 and
g(λ,λ)=g(μ,μ)=0 so

g(λ,μ)=(λ1)(μ1)(λμ)(aλ+bμ+c)

Now expanding and comparing coefficients

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

f=(λ1)(μ1)(λμ)(λ+μ+1)x4 or

f=(λxx)(μxx)(λxμx)(λx+μx+x) or

f=(yx)(zx)(yz)(y+z+x)

Finally

xy(x2y2)+yz(y2z2)+zx(z2x2)=(xy)(xz)(yz)(x+y+z)

Jul 29, 2018

xy(x2y2)+yz(y2z2)+zx(z2x2)

=xy(x2y2)+y3zyz3+z3xzx3

=xy(x2y2)zx3+y3z+z3xyz3

=xy(x2y2)z(x3y3)+z3(xy)

=xy(x+y)(xy)z(xy)(x2+xy+y2)+z3(xy)

=(xy)[xy(x+y)z(x2+xy+y2)+z3]

=(xy)[x2y+xy2zx2xyzy2z+z3]

=(xy)[x2yzx2+xy2xyzy2z+z3]

=(xy)[x2(yz)+xy(yz)z(y2z2)]

=(xy)(yz)[x2+xyzyz2]

=(xy)(yz)[x2z2+xyzy]

=(xy)(yz)[(xz)(x+z)+y(xz)]

=(xy)(yz)(xz)(x+y+z)]