Let's determine the trinomial whose third power is:

$$
x^{6}-6 x^{5}+15 x^{4}-20 x^{3}+15 x^{2}-6 x+1
$$
Given the expression, taking the cube root yields:

$$
x^{2}-2 x+1
$$

However, the sought expression can also be determined as follows: Since the coefficient of the first term is 1 and the constant term is also 1, the sought trinomial is of the form

$$
x^{2}+a x+1
$$

If we cube this trinomial and arrange the resulting expression in descending powers of \( x \), we get:

$$
x^{6}+3 a x^{5}+\left(3 a^{2}+3\right) x^{4}+\left(a^{3}+6 a\right) x^{3}+\left(3+3 a^{2}\right) x^{2}+3 a x+1
$$

The given expression is identically equal to the latter, and thus the corresponding coefficients must also be equal. Therefore, it must be:

$$
3 a=-6, \quad 3 a^{2}+3=15, \quad a^{3}+6 a=-20, \quad 3+3 a^{2}=15, \quad 3 a=-6
$$

From each of these equations, it follows that \( a=-2 \), and thus the sought trinomial is:

$$
x^{2}-2 x+1
$$

(Freund Ernö, Budapest.)

The problem was solved by: Ehrenstein P., Felhőssy J., Frank T., Füstös P., Gerő M., Harsányi Z., Heimlich P., Jánosy Gy., Kirchknopf E., Kiss J., Makó E., Paunz A., Petrik S., Sárközy E., Schlesinger Gy., Schuster Gy., Schwarz Gy., Steiger J., Stolzer J., Szabó J., Szécsi I., Tóth B.