Easy. Deduce $(xy+yz+zx)(x+y+z) = xyz$, and use the observation $(x+y+z)(xy+yz+zx) - (x+y)(y+z)(z+x)=xyz$ to arrive at $(x+y)(y+z)(z+x)=0$. From here, the conclusion is immediate.

$P(X)=(X+x)(X+y)(X+z)=X^3+aX^2+\frac{p}{a}X+p$ where $p=xyz$. $P(-a)=0$. Done

We multiply by $(x+y+z)$: $$\frac{y}{x}+\frac{z}{x}+\frac{x}{y}+\frac{z}{y}+\frac{x}{z}+\frac{y}{z}+3=\frac{x+y+z}{a}$$$$\frac{y+z}{x}+\frac{x+z}{y}+\frac{x+y}{z}=-2$$Substitute $y+z=a-x$, $x+z=a-y$, and $x+y=a-z$ to get $$\frac{a-x}{x}+\frac{a-y}{y}+\frac{a-z}{z}=-2$$Take the case where $a=0$. Then, exactly one of the variables must be one. $$\frac{x-a}{x}+\frac{y-a}{y}+\frac{z-a}{z}=2$$$$\frac{a}{x}+\frac{a}{y}+\frac{a}{z}=1$$If $a<0$, idkkk I'm overcomplicating it

Suppose that $x,y,z\neq a$. Then, we have $x+y=a-z$ and $\frac{x+y}{xy}=\frac{z-a}{az}$, so we have $$\frac{a-z}{xy}=\frac{z-a}{az}.$$Since $z\neq a$, we can divide both sides by $z-a$ to get $$\frac1{xy}=-\frac1{az},$$so $$xy+az=1.$$Similarly, $xz+ay=1$, so $xy+az=xz+ay$. This is equivalent to $(x-a)(y-z)=$. Since $x\neq a$, we must have $y=z$. Similarly, $x=z$, so $x=y=z=\frac a3$. However, $\frac1x+\frac1y+\frac1z=\frac9a\neq\frac1a$, which is a contradiction. Therefore, at least one of the numbers $x,y,z$ is equal to $a$.

samrocksnature wrote: We multiply by $(x+y+z)$: $$\frac{y}{x}+\frac{z}{x}+\frac{x}{y}+\frac{z}{y}+\frac{x}{z}+\frac{y}{z}+3=\frac{x+y+z}{a}$$$$\frac{y+z}{x}+\frac{x+z}{y}+\frac{x+y}{z}=-2$$Substitute $y+z=a-x$, $x+z=a-y$, and $x+y=a-z$ to get $$\frac{a-x}{x}+\frac{a-y}{y}+\frac{a-z}{z}=-2$$Take the case where $a=0$. Then, exactly one of the variables must be one. $$\frac{x-a}{x}+\frac{y-a}{y}+\frac{z-a}{z}=2$$$$\frac{a}{x}+\frac{a}{y}+\frac{a}{z}=1$$If $a<0$, idkkk I'm overcomplicating it Whoa @samrocksnature here

Multiply and factor: $$(xy+yz+zx)(x+y+z)=xyz.$$Factor some more: $$(x+y)(y+z)(z+x)=0.$$Which, after use of the first equation, becomes: $$(x-a)(y-a)(z-a)=0.~\square$$

An equivalent problem: $$xy+yz+zx=a$$$$\frac1{xy}+\frac1{yz}+\frac1{zx}=\frac1a$$Prove that $a\in\{x,y,z\}$.

Note that $(x-a)(y-a)(z-a)=xyz-a(xy+yz+xz)+a^2(x+y+z)-a^3=0+0=0$, hence one of $x,y,z$ is equal to $a$