Prove $x^n(y-z) + y^n(z-x) + z^n(x-y)$ is divisible by $(y-z)(z-x)(x-y)$

Question

I have to show that $$P(x, y, z) = x^n(y-z) + y^n(z-x) + z^n(x-y)$$

is always divisible by $Q(x, y, z) = (y-z)(z-x)(x-y)$ for $n$ greater than $1$ and I have no idea how to proceed.

-> I would try to prove they have common roots but I don't even know what "to be a root" means on a problem like this so I tried to factor it and realized that this might not be the correct approach. Can someone help me ?

Answer 1

Basic idea: If you let $Q(x,y,z) = P(x,y,z+ x)$, then you can verify directly that $Q(x,y,0) = 0$, which means $z$ factors $Q(x,y,z)$. In other words $Q(x,y,z) = zR(x,y,z)$ for some polynomial $R(x,y,z)$. Hence $P(x,y,z) = Q(x,y,z-x) = (z - x)R(x,y,z-x)$. Thus $P(x,y,z)$ has $z- x$ as a factor.

You can do the analogous steps to get the other two factors.

Answer 2

Hint: You can think $P(x,y,z)$ as a polynomial in $y$ , then see that $P(x,z,z)=0$. This means, $(y-z)$ divides $P(x,y,z)$, as we have done for polynomials in one variable. Similarly think about $P(y,y,z)$ and $P(z,y,z)$.

Answer 3

Note that it suffices to show that $P(x,y,z)$ is divisible by each of the three factors $(x-y)$, $(y-z)$, and $(z-x)$. However, justifying this properly requires the theory of Unique Factorization Domains, which is possibly outside the scope of the question.

One alternative path is to proceed by direct factoring. This starts as follows: $$ \begin{align*} Q(x,y,z) & = x^n(y-z) + y^n(z-x) + z^n(x-y) \\ & = xy(x^{n-1}-y^{n-1}) + z(y^n-x^n)+z^n(x-y) \\ & = (x-y)(xy(x^{n-2}+yx^{n-3}+\ldots+y^{n-2}) -z(y^{n-1}+xy^{n-2}+\ldots+x^{n-1})+z^n)\;. \\ \end{align*} $$ The next step is to explicitly factor $(xy(x^{n-2}+yx^{n-3}+\ldots+y^{n-2}) -z(y^{n-1}+xy^{n-2}+\ldots+x^{n-1})+z^n$ as $(y-z)R(x,y,z)$, and so on. Doing this is looks like a very messy exercise, but it ends up with the following nice form: $$ Q(x,y,z) = -(x-y)(y-z)(z-x)\left(\sum_{i+j+k=n-2}x^iy^jz^k\right)\;. $$ Verifying that this multiplies out to $Q(x,y,z)$ is probably more elegant than the above direct factorization, but does require some care.

Answer 4

If at least 2 of the $x,y,z$ are equal then it is obvious. Eitherwise, You can consider $P$ as a polynomial of $x$ first and see that for $x=y,z$ we get $P=0$, so $$P=(x-y)(x-z)Q (1)$$, where $Q$ is a polynomial of $x$ with $z,y$ as parameters. Now consider $P$ as a polynomial of $z$. For $z=y$, we get again $P=0$, The last one due to $(1)$ leads to the fact that $Q(y)=0$, and so $Q=(z-y)R$, if you consider again $Q$ as a polynomial now of $z$. This concludes the proof.