Decompose the following expression into at least two factors, and if it does not decompose, you must prove with mathematical reasoning why it does not decompose (be sure to say with a mathematical reason why it does not decompose)
$$x^2(x-y)+y^2(y-z)+z^2(z-x)$$
If we permute $x$ to $y$, $y$ to $z$ and $z$ to $x$ at the same time, the result of our decomposition will not change at all, and in addition, we can bring the solution to zero by doing so (by substituting the same mentioned expressions) and we can find the root calculated and I got the wrong factors which I think is a bit strange...
Not probably what you want, but you can factor the expression like this:
$$\left(x-\frac{\sqrt[3]{-25 y^3+27 y^2 z+\sqrt{4 \left(-y^2-3 z^2\right)^3+\left(-25 y^3+27 y^2 z+9 y z^2-27 z^3\right)^2}+9 y z^2-27 z^3}}{3 \sqrt[3]{2}}+\frac{\sqrt[3]{2} \left(-y^2-3 z^2\right)}{3 \sqrt[3]{-25 y^3+27 y^2 z+\sqrt{4 \left(-y^2-3 z^2\right)^3+\left(-25 y^3+27 y^2 z+9 y z^2-27 z^3\right)^2}+9 y z^2-27 z^3}}-\frac{y}{3}\right) \left(x+\frac{\left(1-i \sqrt{3}\right) \sqrt[3]{-25 y^3+27 y^2 z+\sqrt{4 \left(-y^2-3 z^2\right)^3+\left(-25 y^3+27 y^2 z+9 y z^2-27 z^3\right)^2}+9 y z^2-27 z^3}}{6 \sqrt[3]{2}}-\frac{\left(1+i \sqrt{3}\right) \left(-y^2-3 z^2\right)}{3\ 2^{2/3} \sqrt[3]{-25 y^3+27 y^2 z+\sqrt{4 \left(-y^2-3 z^2\right)^3+\left(-25 y^3+27 y^2 z+9 y z^2-27 z^3\right)^2}+9 y z^2-27 z^3}}-\frac{y}{3}\right) \left(x+\frac{\left(1+i \sqrt{3}\right) \sqrt[3]{-25 y^3+27 y^2 z+\sqrt{4 \left(-y^2-3 z^2\right)^3+\left(-25 y^3+27 y^2 z+9 y z^2-27 z^3\right)^2}+9 y z^2-27 z^3}}{6 \sqrt[3]{2}}-\frac{\left(1-i \sqrt{3}\right) \left(-y^2-3 z^2\right)}{3\ 2^{2/3} \sqrt[3]{-25 y^3+27 y^2 z+\sqrt{4 \left(-y^2-3 z^2\right)^3+\left(-25 y^3+27 y^2 z+9 y z^2-27 z^3\right)^2}+9 y z^2-27 z^3}}-\frac{y}{3}\right)$$