<s>[INST] 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.
 [/INST] 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 $$A=(y-z)B=(y-z)(x+y)(x+z)$$
</s>