Other ways to factorize $xy\left(x+y\right)+yz\left(y-z\right)-xz\left(x+z\right)$ To factorize $xy\left(x+y\right)+yz\left(y-z\right)-xz\left(x+z\right)$ I used the fact that $x=-y$ and $y=z$ and $x=-z$ make the expression zero. Hence it factorize to $\lambda (x+y)(y-z)(x+z)$ and we can check that the number $\lambda$ is equal to $1$.
I'm looking for other approaches/ideas to factorize the expression.
 A: $xy\left(x+y\right)+yz\left(y-z\right)-xz\left(x+z\right)=xy\left(x+y\right)+yz(y+\color{red}{x}-\color{red}x-z)-xz\left(x+z\right)=xy(x+y)+yz(y+\color{red}x)-yz(\color{red}x+z)-xz(x+z)=y(x+y)(x+z)-z(x+z)(x+y)=(x+y)(x+z)(y-z)$
A: You can observe that, if $x=-y$ or (not "and") $x=-z$ or (not "and") $y=z$ , then the expression equals to zero. Thus, we have
$$xy(x+y)+yz(y-z)-xz(x+z)=A(x + y) (x + z) (y - z)$$
Here, $A$ must be a real number, not an expression. Because, it is enough to see that the coefficient of $x^2y$ equals to $1$. This implies, $A=1$.

Write,
$$A=(x+z)y^2+y(x^2-z^2)-xz(x+z)\\
\Delta_y=(x-z)^2(x+z)^2+4xz(x+z)^2=(x+z)^2(x+z)^2=(x+z)^4$$
Then we have,
$$y_{1,2}=\frac{(z^2-x^2)±(x+z)^2}{2(x+z)}$$
Can you take from here?
A: I think OP approach is the simplest one because it can be done just in mind without paper writing. One of possible another ways
$$A=xy\left(x+y\right)+yz\left(y-z\right)-xz\left(x+z\right)=x^2(y-z)+x(y^2-z^2)+yz(y-z)$$
$(y-z)$ is common factor of $(y-z)$ and $(y^2-z^2)=(y-z)(y+z)$, so we can factor it out:
$$A=(y-z)(x^2+x(y+z)+yz)$$
Then we need only factorize $$B=x^2+x(y+z)+yz$$
If we don't see that
$$B=x^2+xy+xz+yz=x(x+y)+z(x+y)=(x+z)(x+y)$$
we can use standard way of factorizing quadratic expression with full square separation
$$B=x^2+x(y+z)+yz=x^2+2x\cdot\frac{y+z}2+\frac{(y+z)^2}4-\frac{(y+z)^2-4yz}4$$
$$B=\left(x+\frac{y+z}2\right)^2-\frac{y^2-2yz+z^2}4=\left(x+\frac{y+z}2\right)^2-\left(\frac{y-z}2\right)^2$$
$$B=\left(x+\frac{y+z}2+\frac{y-z}2\right)\left(x+\frac{y+z}2-\frac{y-z}2\right)=(x+y)(x+z)$$
Then $$A=(y-z)B=(y-z)(x+y)(x+z)$$
A: This method can be faster.

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*Write the expression as a general quadatic respect to $y$ ( or $x$, or $z$)



You get,
$$A=(x+z)y^2+y(\color {red}{x^2-z^2})-xz(x+z)=(x+z)(y^2-y(z-x)-zx)$$

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*Apply the Vieta's formulas
$z$ and $-x$ are the root of the quadatic: $$y^2-y(z+(-x))+z\times (-x)$$
Hence we have
$$A=(x+z)(y+x)(y-z).$$
