Other ways to factorize $xy\left(x+y\right)+yz\left(y-z\right)-xz\left(x+z\right)$

Question

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.

Answer 1

$xy\left(x+y\right)+yz\left(y-z\right)-xz\left(x+z\right)=xy\left(x+y\right)+yz(y+\color{red}{x}-\color{red}x-z)-xz\left(x+z\right)=xy(x+y)+yz(y+\color{red}x)-yz(\color{red}x+z)-xz(x+z)=y(x+y)(x+z)-z(x+z)(x+y)=(x+y)(x+z)(y-z)$

Answer 2

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, $A=1$.

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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?

Answer 3

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 $$A=(y-z)B=(y-z)(x+y)(x+z)$$

Answer 4

This method can be faster.

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

You get,

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

• 2. 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

$$A=(x+z)(y+x)(y-z).$$