Find such numbers $A, B, C, a, b, c$, so that the identity

$$
(4 x-2) /\left(x^{3}-x\right)=A /(x-a)+B /(x-b)+C /(x-c)
$$

holds.
$$
(4 x-2) /\left(x^{3}-x\right)=(A(x-b)(x-c)+B(x-a)(x-c)+C(x-a)(x-b)) /((x-a)(x-b)(x-c))
$$

Since both fractions are irreducible, they are equal under the condition of equality of their numerators and denominators:

$$
x^{3}-x=x(x-1)(x+1)=(x-a)(x-b)(x-c)
$$

From this, it is clear that we can set $a=0, b=1, c=-1$. Let's write the condition for the equality of the numerators, taking into account the found values of $a, b, c$:

$$
4 x-2=A(x-1)(x+1)+B(x+1) x+C x(x-1)
$$

or

$$
4 x-2=(A+B+C) x^{2}+(B-C) x-A
$$

From this, we obtain three necessary equations to determine $A, B, C: A+B+C=0, B-C=4, A=2$. Solving them, we find: $A=2, B=1, C=-3$. Therefore,

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

## Answer

$A=2, B=1, C=-3, a=0, b=1, c=-1$.