To show that x^5 - 5x^3 + 5x^2 - 1 = 0 has three equal roots, we can use the fact that if a polynomial has a repeated root of multiplicity n, then the derivative of the polynomial also has that root of multiplicity n-1.
First, we can find the derivative of the polynomial:
f(x) = x^5 - 5x^3 + 5x^2 - 1
f'(x) = 5x^4 - 15x^2 + 10x
Next, we can set f(x) equal to zero and solve for x:
x^5 - 5x^3 + 5x^2 - 1 = 0
We can then substitute this value of x into the derivative f'(x):
f'(x) = 5x^4 - 15x^2 + 10x
f'(x) = 5(x^4 - 3x^2 + 2)
f'(x) = 5(x^2 - 1)(x^2 - 2)
We can see that the derivative has two roots, x = -1 and x = 1, each with multiplicity 2. This implies that the original polynomial has three equal roots, since each of these roots will be a repeated root of multiplicity 2.
To find the roots, we can use synthetic division to divide the polynomial by (x - 1)^2:
1 | 1 0 -5 5 -1
| 1 -1 4 -1
| 1 1 -6 9 -2
1 | 1 1 -6 9 -2
| 1 -5 1
| 1 2 -11 10
So, (x - 1)^2 is a factor of the polynomial, and we can factor the remaining quadratic as:
(x - 1)^2(x^2 + 2x - 10) = 0
Solving for the roots of the quadratic factor, we get:
x^2 + 2x - 10 = 0
(x + 1 + sqrt(11))(x + 1 - sqrt(11)) = 0
Therefore, the roots of the original polynomial are:
x = 1 (with multiplicity 2)
x = -1 + sqrt(11)
x = -1 - sqrt(11)
So, the three equal roots of the polynomial are 1, 1, and -1 ± sqrt(11).