Verify if the fraction (x-1)(x^3+1)x/(x^3-x) >0, if x is a real number?

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.