Factor the following polynomial into the product of two integer-coefficient third-degree polynomials:

$$
x^{6}-x^{5}+x^{4}-x^{3}-x^{2}-x+1
$$
We need to find integer coefficients $A, B, C, D$ and $a, b, c, d$ such that

$$
x^{6}-x^{5}+x^{4}-x^{3}-x^{2}-x+1=\left(A x^{3}+B x^{2}+C x+D\right)\left(a x^{3}+b x^{2}+c x+d\right)
$$

is an identity, meaning that the coefficients of the powers of $x$ on both sides are equal. From the coefficient of $x^{6}$, we have $A a=1$, which is only possible if $A=a=1$ or $A=a=-1$. The latter, however, can be reduced to the former by multiplying both factor polynomials by $-1$, which is not a significant transformation, so it can be ignored. Let us assume $A=a=1$. Then, from the coefficients of $x^{5}, x^{4}, \ldots, x$ and the constant term, we get the following system of equations for the coefficients:

$$
\begin{aligned}
B+b & =-1 \\
C+B b+c & =1 \\
D+C b+B c+d & =-1 \\
D b+C c+B d & =-1 \\
D c+C d & =-1 \\
D d & =1
\end{aligned}
$$

From the last equation, $D=d= \pm 1$, so $d^{2}=1$; based on this, multiplying (5) by $d$ gives

$$
C+c=-d
$$

and from (2),

$$
B b=1+d
$$

Subtract twice (2a) from the square of (1):

$$
B^{2}+b^{2}=-1-2 d
$$

This is non-negative only for $d=-1$. In this case, one of $B$ and $b$ is 0, and the other is -1 according to (1). Since $A=a, D=d$, swapping $b$ and $B$ only means swapping the two factors, so we can assume $b=0, B=-1$. Using these values, from (3) we get $c=-1$ and from (5a) $C=2$. With these values, (4) is also satisfied, so the desired factorization is

$$
\left(x^{3}-x^{2}+2 x-1\right)\left(x^{3}-x-1\right)
$$

Próhle Tamás (Budapest, Fazekas M. Gyak. Gimn., III. o. t.)