We will make use of the fact that if P(x) is a polynomial, then (x-a) is a factor of P(x) if and only if P(a) = 0, along with polynomial long division.
If we look for a root of x^4-2x^3+2x-1 (that is, a value such that the expression evaluates to 0), we find that 1^4-2(1)^3+2(1)-1=0. Thus, using the above fact, we know that (x-1) is a factor of x^4-2x^3+2x-1. Dividing, we find:
(x^4-2x^3+2x-1)/(x-1) = x^3-x^2-x+1
=> x^4-2x^3+2x-1=(x-1)(x^3-x^2-x+1)
Next, we will look to factor x^3-x^2-x+1. While there are more general ways of factoring cubic expressions, we can still test for roots first. Doing so, we find that 1 is still a root of the remaining cubic equation, and so we once again divide by (x-1):
(x^3-x^2-x+1)/(x-1) = x^2-1
=> (x^3-x^2-x+1) = (x-1)(x^2-1)
For the quadratic equation x^2-1, there are multiple ways of continuing. The easiest is to note that it is an example of the special product (a+b)(a-b) = a^2-b^2 with a=x and b=1. We could also test for roots as we did above and divide. Either way, we will find that
x^2-1 = (x+1)(x-1)
Now that we have factored everything down to polynomials of degree 1, we can put it all together to get our factored form:
x^4-2x^3+2x-1 = (x-1)(x^3-x^2-x+1)
=(x-1)(x-1)(x^2-1)
=(x-1)(x-1)(x+1)(x-1)
=(x-1)^3(x+1)
