Calculate the value of the polynomial $P(x)=x^{3}-3 x^{2}-3 x-1$ at the following points:

$$
\begin{gathered}
x_{1}=1-\sqrt{2}, \quad x_{2}=1+\sqrt{2}, \quad x_{3}=1-2 \sqrt{2}, \quad x_{4}=1+2 \sqrt{2}, \\
x_{5}=1+\sqrt[3]{2}+\sqrt[3]{4}
\end{gathered}
$$

and show that $P\left(x_{1}\right)+P\left(x_{2}\right)=P\left(x_{3}\right)+P\left(x_{4}\right)$.
Each substitution becomes easier if we rearrange the polynomial in powers of $x-1$, determining the new coefficients. The

$$
P(x) \equiv Q(x-1)=(x-1)^{3}+a(x-1)^{2}+b(x-1)+c
$$

requirement must hold for all $x$. Choosing the value of $x$ as 1, 0, and 2, we get a simple system of equations for $a, b, c$:

$$
\begin{array}{llr}
x=1, & P(1)=Q(0), & -6=c \\
x=0, & -1=-1+a-b+c \\
x=2, & P(0)=Q(-1), & -11=1+a+b+c
\end{array}
$$

from which $c=-6, a=0, b=-6$, so

$$
P(x)=x^{3}-3 x^{2}-3 x-1 \equiv(x-1)^{3}-6(x-1)-6 \text {. }
$$

Now,

$$
\begin{aligned}
& P\left(x_{1}\right)=P(1-\sqrt{2})=Q(-\sqrt{2})=-2 \sqrt{2}+6 \sqrt{2}-6=4 \sqrt{2}-6 \\
& P\left(x_{2}\right)=Q(\sqrt{2})=-4 \sqrt{2}-6 ; \quad P\left(x_{1}\right)+P\left(x_{2}\right)=-12 \\
& P\left(x_{3}\right)=Q(-2 \sqrt{2})-4 \sqrt{2}-6, \quad P\left(x_{4}\right)=Q(2 \sqrt{2})=4 \sqrt{2}-6 \\
& P\left(x_{3}\right)+P\left(x_{4}\right)=-12
\end{aligned}
$$

the two sums are indeed equal. - Finally,

$$
\begin{aligned}
P\left(x_{5}\right)= & Q(\sqrt[3]{2}+\sqrt[3]{4})=(\sqrt[3]{2}+\sqrt[3]{4})^{3}-6(\sqrt[3]{2}+\sqrt[3]{4})-6= \\
& 2+6 \sqrt[3]{2}+6 \sqrt[3]{4}+4-6 \sqrt[3]{2}-6 \sqrt[3]{4}-6=0
\end{aligned}
$$

so $x_{5}$ is a root of the equation $P(x)=0$.

Péter Békéssy (Budapest, Berzsenyi D. g. I. o. t.)

Note. The equality of the two sums holds for any parameter $a$ defining $x_{1}=1-a, x_{2}=1+a, x_{3}=1-2 a$, $x_{4}=1+2 a$.

Judit Bod (Budapest, Apáczai Csere J. gyak. g. III. o. t.)