We have to determine if the fraction (x-1)(x^3+1)x/(x^3-x) is always greater than 0 or not.

(x-1)(x^3+1)x/(x^3-x)

=> x(x - 1)(x + 1)(x^2 - x + 1)/x*(x - 1)(x + 1)

=> x^2 - x + 1

=> x^2 - x + 1/4 + 3/4

=> (x - 1/2)^2 + 3/4

As the square of a real number is always positive (x - 1/2)^2 + 3/4 is also positive.

**This proves that (x-1)(x^3+1)x/(x^3-x) > 0 if x is real.**

First, we'll factorize the denominator by x:

(x-1)(x^3+1)x/x(x^2-1)

We'll reduce by x:

(x-1)(x^3+1)/(x^2-1)

Since x^3 + 1 is a sum of cubes, then it could be written as it follows:

x^3 + 1 = (x+1)(x^2 - x + 1)

Since x^2 - 1 is a difference of squares, then it could be written as it follows:

x^2 - 1 = (x-1)(x+1)

We'll re-write the fraction:

(x-1)(x^3+1)/(x^2-1) = (x-1)(x+1)(x^2 - x + 1)/(x-1)(x+1)

We'll simplify by (x-1)(x+1):

(x-1)(x^3+1)/(x^2-1) = x^2 - x + 1

We'll verify if the quadratic has any zeroes:

x^2 - x + 1 = 0

delta = (-1)^2 - 4 = 1 - 4 = -3 < 0

Since the discriminant is negative, then the quadratic is not intercepting x axis and it is located above or below x axis. Since the coefficients of x^2 is positive, then the quadratic x^2 - x + 1 is positive for any value of x.

**Therefore, the statement that the fraction (x-1)(x^3+1)x/x(x^2-1) is positive for any value of x is true.**