Solve the following equation if $t$ is a given real number:

$$
x^{3}-2 t x^{2}+t^{3}=0
$$
Factorize the left side of the equation:

$$
x^{3}-t x^{2}-t x^{2}+t^{3}=x^{2}(x-t)-t\left(x^{2}-t^{2}\right)=x^{2}(x-t)-t(x-t)(x+t)=(x-t)\left(x^{2}-t(x+t)\right)=0
$$

A product is zero if and only if one of its factors is zero.

Thus, either $x-t=0$, which means $x=t$, or $\left(x^{2}-t(x+t)\right)=0$, which is a quadratic equation in $x$ with roots:

$$
x_{1}=\frac{t+t \sqrt{5}}{2}, \quad x_{2}=\frac{t-t \sqrt{5}}{2}
$$

Therefore, the solutions to the equation are

$$
t, \quad t \frac{1+\sqrt{5}}{2}, \quad t \frac{1-\sqrt{5}}{2}
$$

Kálmán Barnabás (Budapest, ELTE Apáczai Csere J. Gyak. Gimn., IV. o.t.)