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(1) $ z^2y+xy^2+x^2z-(x^2y+xz^2+y^2z) $
(2) $ (x-y)(y-z)(z-x) $

How to go from STEP (1) to STEP (2). Nothing I do seems to work. I tried combining terms but that doesn't help.

I do not want to go from step 2 to step 1. I arrived at step 1 in some question and I need to go from 1 to 2 to match my answer given in the textbook.

I see a lot of comments asking me to just expand 2 and arrive at 1. I could see that too but I am really curious to know how it is done the other way around. Added to that, if you get (1) while solving some question, you obviously have to go from 1 to 2 and not from 2 to 1.

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    Is it forbidden to just multiply out $(2)$ and the compare it to $(1)$? I don't see the sense of using some "trick". – 2017-02-17
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    Just expanding works fine. – 2017-02-17
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    Note that if $x=y$ the first quantity would be zero ... this means $(x-y)$ is a factor. Similarly $(y-z)$ and $(z-x)$ are factors. So (1) = $k(x-y)(y-z)(z-x)$ ... just need to find $k$ now ! – 2017-02-17
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    @Cotton You can check out my answer. I think that is the method you are looking for. – 2017-02-17

4 Answers 4

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$\begin{array}\\ z^2y+xy^2+x^2z-(x^2y+xz^2+y^2z) &=z^2y-z^2x+xy^2-x^2y+x^2z-y^2z\\ &=z^2(y-x)+xy(y-x)+z(x^2-y^2)\\ &=z^2(y-x)+xy(y-x)+z(x+y)(x-y)\\ &=(y-x)(z^2+xy-z(x+y))\\ &=(y-x)(z^2-zx+xy-zy)\\ &=(y-x)(z(z-x)+y(x-z))\\ &=(y-x)(z-y)(z-x)\\ \end{array} $

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Another method :

Rewrite (1) as

$$-yx^2 + x^2 z + x y^2 - x z^2 - y^2 z + y z^2 + \underline{ xyz - xyz}$$

Group the terms as follows :

$$(\underline{xyz - xz^2 - y^2z + yz^2} ) - ( \underline{ x^2y-x^2z-xy^2+xyz})$$

Factor out $z$ from first underlined expression and $x$ from the second :

$$ = z(xy - xz -y^2 + yz) - x(xy - xz - y^2 + yz)$$

Now it can be easily seen that the two expressions inside the brackets are identical, so factor them out :

$$=(z-x)(xy - xz -y^2 + yz)$$

$$=(z-x)(x(y-z)-y(y-z))$$

$$=(z-x)(y-z)(x-y)$$

As Desired.

So the "trick" here is to add and subtract $abc$. It can actually be observed quite easily by multiplying the brackets in $(2)$.

Hope this helps. :-)

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Pair the positive and negative terms to get one of the factors \begin{eqnarray*} z^2y+xy^2+x^2z-(x^2y+xz^2+y^2z) = z^2y -xz^2+xy^2-x^2y +x^2z-y^2z \end{eqnarray*} Pull off this factor \begin{eqnarray*} (x-y)(-z^2-xy+(x+y)z) \end{eqnarray*} The content of the second bracket is easily shown to be what is required.

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Hint: the first expression can easily seen to be $0$ for $x=y$ and $x=z$. Considering it as a polynomial in $x\,$ of degree $2$, this means that it factors as $\lambda(x-y)(x-z)$ where the dominant coefficient $\lambda$ must match the coefficient of $x^2$ from the original expression i.e. $\lambda=z-y$.