So long as the denominator isn't 0 for a particular value of x this equation may as well read

x^3 - x^2 - x + 1 = 0

(Just multiply both sides by x^3 - x^2 + x - 1)

Just for reference when does x^3 - x^2 + x - 1 = 0?

Let's apply the Rational Root theorem:

Given a polynomial equation p*x^n + a*x^{n-1} + ... + bx + q = 0, the rational roots (if any exist) take the form of x = (+/-){factor of q}/{factor of p}.

In this case p = 1 and q = -1. So the rational roots of x^3 - x^2 + x - 1 = 0, if any exist, take the form of x = (+/-)1. We see immediately that x = 1 is a root. Thus x^3 - x^2 + x - 1 = (x - 1)(x^2 + 1) = 0 (by long division.) So the other two roots are x = (+/-)I, where I^2 = -1.

Thus our problem is

x^3 - x^2 - x + 1 = 0, x not equal to -1, (+/-)I.

Again, apply the rational root theorem.

Any roots of this equation will take the form of x = (+/-)1. We see that x = 1 is again a root. Thus

x^3 - x^2 - x + 1 = (x - 1)(x^2 - 1) = 0 (by long division)

The remaining two solutions are thus x = (+/-)1.

So our solution set is x = {-1, 1}, BUT we can't have x = -1. Thus the only solution to

x^3-x^2-x+1

----------------- = 0

x^3-x^2+x-1

is x = 1.

-Dan