Solve for $x, y, z \in \mathbb R$: $x[x^2 + (y - z)^2] = 2, y [y^2 + (z - x)^2] = 16, z[z^2 + (x - y)^2] = 30$.

Question

Solve the following system of equations $$\large \left\{ \begin{align} x \cdot \left[x^2 + (y - z)^2\right] &= 2\\ y \cdot \left[y^2 + (z - x)^2\right] &= 16\\ z \cdot \left[z^2 + (x - y)^2\right] &= 30 \end{align} \right.$$

My answer is provided below. The first step is always to prove $2y = z + x$. After that, you can do whatever you want to further find the relations between $x$, $y$ and $z$(, at least I think so).

Answer 1

$(x, y, z > 0)$

We have that $$z \cdot \left[z^2 + (x - y)^2\right] + x \cdot \left[x^2 + (y - z)^2\right] - 2y \cdot \left[y^2 + (z - x)^2\right] = 0$$

$$ \iff z \cdot \left(z^2 + x^2 + y^2 - 2xy\right) + x \cdot \left(x^2 + y^2 + z^2 - 2yz\right) - 2y \cdot \left(y^2 + z^2 + x^2 - 2zx\right) = 0$$

$$ \iff (z + x - 2y)(y^2 + z^2 + x^2) = 0 \implies z + x - 2y = 0 \iff \frac{z + x}{2} = y \tag 1$$

Plugging in $\dfrac{z + x}{2} = y$, we have that

$$\left\{ \begin{align} x \cdot \left[4x^2 + (x - z)^2\right] &= 8\\ z \cdot \left[4z^2 + (x - z)^2\right] &= 120 \end{align} \right.$$

$$ \iff \left\{ \begin{align} x(5x^2 - 2zx + z^2) &= 8\\ z(5z^2 - 2zx + x^2) &= 120 \end{align} \right.$$

$$\implies z(5z^2 - 2zx + x^2) - 15x(5x^2 - 2zx + z^2) = 0$$

$$ \iff (z - 3x)(5z^2 - 2zx + 25x^2) = 0 \iff z = 3x \tag 2$$

From $(1)$ and $(2)$, we have that $x = \dfrac{y}{2} = \dfrac{z}{3} \implies \left\{ \begin{align} x - y &= -\frac{z}{3}\\ y - z &= -x\\ z - x &= y \end{align} \right.$

Plugging the previous statement into the equations, we have that $(x, y, z) = (1, 2, 3)$.

Answer 2

We have $$x^3+x(y-z)^2+y^3+y(x-z)^2-2(z^3+z(x-y)^2)=0$$ or $$(x+y-2z)(x^2+y^2+z^2)=0.$$ Can you end it now?