I'm trying to factorise $$ x^3z - x^3y - y^3z + yz^3 + xy^3 - xz^3 $$ into four linear factors. By plugging it into WolframAlpha I've learned that it's $$-(x-y)(x-z)(y-z)(x+y+z)$$

My question is: what are the steps involved in factorising the expression? Is there a method I don't know about that I'd have access to with my limited maths?

Really appreciate any help!

  • 3
    $\begingroup$ FWIW, I think it looks prettier as $(x-y)(y-z)(z-x)(x+y+z)$, that way you see the cyclic nature of the three binomials and there is no arbitrary $-$ sign. $\endgroup$
    – MCT
    Feb 20, 2015 at 23:01
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    $\begingroup$ Try $y=x$ first and you'll get $0$ so... $\endgroup$
    – mathlove
    Feb 20, 2015 at 23:02
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    $\begingroup$ Basically, just observe that it is reasonably symmetric in $x$, $y$ and $z$. "Eye-balling it" we can see that it is zero for $x = y$, $y = z$ and $z = x$. You can then write it as $(x-y)(y-z)(z-x)(Ax + By + Cz)$ (you can see that the orders are right), expand this and solve for $A$, $B$ and $C$; alternatively, you could do long division. $\endgroup$
    – Sam OT
    Feb 20, 2015 at 23:06
  • $\begingroup$ There are factoring techniques that generalize what you are being encouraged to "eyeball". One approach is to substitute values for two of the three variables and consider the resulting factors with respect to the third variable. One can then "piece together" some plausible guess at what those factors are in terms of the original three variables. Would an answer along these lines (factoring multivariate polynomials) be of interest? $\endgroup$
    – hardmath
    Feb 20, 2015 at 23:10
  • $\begingroup$ Anything like that would be great. So far my approach was taking out $x$ from two terms, $y$ from two terms and $z$ from two terms which immediately hit a dead end. I hadn't heard of setting $x = y$, $y = z$, $z = x$, so thanks, I'll try that. $\endgroup$
    – cge
    Feb 20, 2015 at 23:15

4 Answers 4


mathlove has a great hint. The way to factor polynomials is to find their zeroes. That is to say, something like $x^2+2x-3$ can be factored as $(x-1)(x+3)$, where $x=1$ and $x=-3$ are the zeroes. So, playing around with this equation you may think about trying something like $x=y$ to see what happens. This is great, because it causes the whole expression to equal zero, so you know that $x=y$ is a zero. This means you can factor out a quantity of $(x-y)$. Doing polynomial division you will find $$ x^3z - x^3y - y^3z + yz^3 + xy^3 - xz^3 = (x-y)(x^2z+xyz+y^2z-x^2y-xy^2-z^3)$$ Now you can repeat this process on the quantity $x^2z+xyz+y^2z-x^2y-xy^2-z^3$ and factor out another zero, proceeding until everything has been factored. You already know which quantities can be factored out so it shouldn't be too bad. But going through polynomial division is great practice, and is often necessary for factoring polynomials of degree three or higher.

  • $\begingroup$ Thanks! Yeah, I was missing two things: that factoring should be seen as finding roots and that if the expression is 0 when e.g. $y = x$ then $x - y$ is a root. The hint from @mathlove was very useful. $\endgroup$
    – cge
    Feb 20, 2015 at 23:33
  • $\begingroup$ In fact, since it's clear even from eyeballing that the original expression is cyclically symmetric in $(x,y,z)$ then once you know that $(x-y)$ is a factor you immediately know that $(y-z)$ and $(z-x)$ are too. $\endgroup$ Feb 20, 2015 at 23:54
  • $\begingroup$ @StevenStadnicki I hadn't thought of it that way. Then again, I am not familiar with cyclically symemetric polynomials, and am still a "guess-and-checker." Always nice when multiple perspectives point to the same answer though! $\endgroup$
    – graydad
    Feb 20, 2015 at 23:58

Let's do this inelegantly, just to show that it can be done.

First write as a polynomial (cubic) in $x$ to give $$(z-y)x^3+(y^3-z^3)x+yz^3-y^3z=(z-y)x^3-(z-y)(z^2+yz+y^2)x+yz(z-y)(y+z)=$$$$=(z-y)(x^3-z^2x-xyz-y^2x+y^2z+yz^2)$$

Now tackle the second bracket as a quadratic in $y$ viz $$(z-x)y^2+z(z-x)y+x(x+z)(x-z)=(z-x)(y^2+zy-x^2-xz)$$(we could have used the quadratic formula here to solve for $y$)

Now the second bracket here is linear in $z$ $$(y-x)z+(y-x)(y+x)=(y-x)(z+y+x)$$

Put the pieces together to get $(z-y)(z-x)(y-x)(x+y+z)$

  • $\begingroup$ Is there a reason for the stray $=$ in the middle there? $\endgroup$ Feb 20, 2015 at 23:32
  • $\begingroup$ @crash I sometimes do this so that a person reading the top line will see that something comes afterwards, and someone reading the bottom line will see that something came before. It's a personal peculiarity, I guess. $\endgroup$ Feb 20, 2015 at 23:39
  • $\begingroup$ Interesting. I've never seen that done before. I personally prefer using the align environment to make all steps or deductions as clear as possible, tagging anything that might need a minor comment on the side. To each his own I guess :) $\endgroup$ Feb 20, 2015 at 23:42
  • $\begingroup$ @MarkBennet I never thought this could be called a "peculiarity" and I thought it was common practice. All my life I've always did this with all arithmetical operations and everything else, not only with $=$. $\endgroup$
    – user26486
    Feb 20, 2015 at 23:53

Here's the way we did this when I was beginning high school: as it's a symmetric function of 3 variables, at some point we have to break the symmetry. All we meed is remarkable identities:

The expression can be rewritten as \begin{align*} (x^3-y^3)z&+(z^3-x^3)y+(y^3-z^3)x =(x^3-y^3)z+(z^3-y^3+y^3-x^3)y+(y^3-z^3)x\\ & =(x^3-y^3)(z-y)+(y^3-z^3)(x-y)\\ & =(x-y)(z-y)(x^2+xy+y^2)+(x-y)(y-z)(y^2+yz+z^2)\\ &= (x-y)(y-z)(z^2+yz-x^2-xy)\\ &= (x-y)(y-z)\bigl((z-x)(z+x)+y(z-x)\bigr)\\ &=(x-y)(y-z)(z-x)(x+y+z). \end{align*}


Because the polynomial is homogeneous in degree (every term is 4th degree), the expression may reduced to two variables by "factoring out" a 4th power of one of the variables.

If we set $X=x/z$ and $Y=y/z$, then the original expression is $z^4$ times the following:

$$ X^3 - X^3Y - Y^3 + Y + XY^3 - X $$

Collecting terms in descending powers of one variable (say $X$), we have:

$$ (1-Y)X^3 + (Y^3 - 1)X - Y^3 + Y $$

By inspection the leading coefficient is a factor:

$$ (1-Y)(X^3 - (Y^2 + Y + 1)X + (Y + 1)Y) $$

Since the second factor is now monic in $X$ and polynomials in $X$ over $\mathbb{Z}[Y]$ form a UFD, we check for factors of the form $X-d$ where root $d$ is a divisor of $(Y+1)Y$. Equivalently we can substitute $X=d$ rather than perform long division, which amounts to synthetic division/Horner's method if you practice it. In any case one gets the factor $X-Y$ rather easily:

$$ (1-Y)(X-Y)(X^2 + YX - (Y+1)) $$

Finally the third factor yields either to inspection, to root testing, or to the quadratic formula:

$$ (1-Y)(X-Y)(X - 1)(X + Y + 1) $$

Multiplying by $z^4$ in such a way that each factor above gets one copy of $z$ converts back to the original variables:

$$ (z-y)(x-y)(x-z)(x + y + z)$$

And we are done!


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