Provide a fifth-degree polynomial $f$ such that $(x-1)^{3}$ can be factored out of $f(x)+1$ and $(x+1)^{3}$ can be factored out of $f(x)-1$.
I. solution. If $f(x)$ is a fifth-degree polynomial, then $f(x)+1$ and $f(x)-1$ are also fifth-degree polynomials. If $(f(x)+1)$ can be factored by $(x-1)^{3}$, then there exists a second-degree polynomial $q(x)=a x^{2}+b x+c$ such that

$$
f(x)+1 \equiv(x-1)^{3}\left(a x^{2}+b x+c\right)
$$

and similarly, if $(f(x)-1)$ can be factored by $(x+1)^{3}$, then there exists a second-degree polynomial $p(x)=d x^{2}+e x+g$ such that

$$
f(x)-1 \equiv(x+1)^{3}\left(d x^{2}+e x+g\right)
$$

Subtracting (2) from (1):

$$
2 \equiv(x-1)^{3}\left(a x^{2}+b x+c\right)-(x+1)^{3}\left(d x^{2}+e x+g\right)
$$

or

$$
\begin{aligned}
2 \equiv & (a-d) x^{5}+(b-3 a-e-3 d) x^{4}+(c-3 b+3 a-g-3 e-3 d) x^{3}+ \\
& +(-3 c+3 b-a-3 g-3 e-d) x^{2}+(3 c-b-3 g-e) x-c-g
\end{aligned}
$$

The identity holds for all $x$ only if the coefficients of the higher-degree terms on the right side are zero, and

$$
-(c+g)=2
$$

and the coefficients of the higher-degree terms are zero, so

$$
\begin{gathered}
a=d, \quad b-3 a=e+3 d, \quad c-3 b+3 a=g+3 e+3 d \\
-3 c+3 b-a=3 g+3 e+d, \quad 3 c-b=3 g+e
\end{gathered}
$$

The system of linear equations (3) and (4) consists of 6 unknowns and 6 equations. Solving it using any known method:

$$
a=d=-\frac{3}{8}, \quad b=-\frac{9}{8}, \quad e=\frac{9}{8}, \quad c=g=-1
$$

From (1) for the desired polynomial:

$$
f(x)=(x-1)^{3}\left(-\frac{3}{8} x^{2}-\frac{9}{8} x-1\right)-1=-\frac{3}{8} x^{5}+\frac{5}{4} x^{3}-\frac{15}{8} x
$$

This polynomial is indeed a solution, and it is clear from our solution that it is the only such polynomial.

II. solution. If $(f(x)+1)$ can be factored by $(x-1)^{3}$, then its derivative, $f^{\prime}(x)$, is divisible by $(x-1)^{2}$. If $(f(x)-1)$ can be factored by $(x+1)^{3}$, then the derivative of $(f(x)-1)$, $f^{\prime}(x)$, can also be factored by $(x+1)^{2}$. Thus, we get that $f^{\prime}(x)$ can be factored by

$$
(x-1)^{2}(x+1)^{2}=\left(x^{2}-1\right)^{2}=x^{4}-2 x^{2}+1
$$

Since $f(x)$ is a fifth-degree polynomial, $f^{\prime}(x)$ is a fourth-degree polynomial. Therefore,

$$
f^{\prime}(x)=A\left(x^{4}-2 x^{2}+1\right)
$$

Consequently,

$$
f(x)=\frac{A x^{5}}{5}-\frac{2 A x^{3}}{3}+A x+C
$$

To determine $A$ and $C$, we need to consider that $f(x)+1$ is divisible by $(x-1)^{3}$, so $f(1)+1=0$, i.e., $f(1)=-1$.

$$
-1=f(1)=\frac{A}{5}-\frac{2 A}{3}+A+C
$$

On the other hand, $f(x)-1$ is divisible by $(x+1)^{3}$, so $f(-1)-1=0$, i.e., $f(-1)=1$.

$$
1=f(-1)=-\frac{A}{5}+\frac{2 A}{3}-A+C
$$

Adding (5) and (6) gives $0=2 C$, so $C=0$. Subtracting (5) from (6) gives $-\frac{16}{15} A=2$, so $A=\frac{-15}{8}$. Therefore,

$$
f(x)=-\frac{3}{8} x^{5}+\frac{5}{4} x^{3}-\frac{15}{8} x
$$

This polynomial (and only this one) satisfies the conditions.

III. solution. If $f(x)-1$ can be factored by $(x+1)^{3}$, then $f(-x)-1$ can be factored by $(-x+1)^{3}$, i.e., $-(x-1)^{3}$. Thus, $(x-1)^{3}$ can be factored from $f(x)+1$ and $f(-x)-1$, and therefore from their sum, $f(x)+f(-x)$. Similarly, we get that $(x+1)^{3}$ can also be factored from $f(x)+f(-x)$.

The polynomial $f(x)+f(-x)$ is also (at most) a fifth-degree polynomial, and $(x+1)^{3}$ and $(x-1)^{3}$ can be factored from it, so $(x+1)^{3}(x-1)^{3}$ can also be factored from it (since the polynomials $(x+1)^{3}$ and $(x-1)^{3}$ have no common polynomial divisor). A sixth-degree polynomial can only be factored from a (at most) fifth-degree polynomial if the polynomial is identically zero. Therefore,

$$
f(x)+f(-x) \equiv 0, \quad \text { i.e., } \quad f(x) \equiv-f(-x)
$$

i.e., $f(x)$ is an odd function.

Let $f(x)=a_{5} x^{5}+a_{4} x^{4}+a_{3} x^{3}+a_{2} x^{2}+a_{1} x+a_{0}$. According to (7), $f(0)=-f(0)$, so $a_{0}=f(0)=0$. The function $a_{5} x^{5} a_{3} x^{3}+a_{1} x$ is odd, and the difference of odd functions is also odd, so the function

$$
g(x)=a_{4} x^{4}+a_{2} x^{2}=f(x)-a_{5} x^{5}-a_{3} x^{3}-a_{1} x
$$

is odd. On the other hand, $g(x)$ is clearly even, so $g(-x) \equiv g(x) \equiv-g(-x)$, and thus $g(x) \equiv 0$. Therefore, $g(x)$ is identically zero, and thus $a_{4}=a_{2}=0$, so $f(x) \equiv a_{5} x^{5}+a_{3} x^{3}+a_{1} x$.

$(f(x)+1)$ can be factored by $(x-1)^{3}$, so its derivative, $f^{\prime}(x)$, can be factored by $(x-1)^{2}$, and $f^{\prime \prime}(x)$ can be factored by $(x-1)$. Therefore, $f(1)+1=$ $f^{\prime}(1)=f^{\prime \prime}(1)=0$. Substituting these into the

$$
\begin{aligned}
& 0=f(1)=a_{5}+a_{3}+a_{1}+1 \\
& 0=f^{\prime}(1)=5 a_{5}+3 a_{3}+a_{1} \\
& 0=f^{\prime \prime}(1)=20 a_{5}+6 a_{3}
\end{aligned}
$$

system of equations, we get that the solution to this system of equations is $a_{5}=-\frac{3}{8}, a_{3}=\frac{5}{4}, a_{1}=-\frac{15}{8}$. Therefore, the only polynomial that satisfies the problem is $f(x)=-\frac{3}{8} x^{5}+\frac{5}{4} x^{3}-\frac{15}{8} x$, and this polynomial does indeed satisfy the conditions.

Remark. The reasoning in the third solution can be used to show that if the polynomial function $p(x)$ is an odd function, then the coefficients of the even-degree terms are zero. If $p(x)$ is an even function, then the coefficients of the odd-degree terms are 0.