The sum of the three roots of the given polynomial could be found using Viete's relation:

x1 + x2 + x3 = -b/a, where x1,x2,x3 are the roots of the polynomial and a,b are the coefficients of the polynomial ax^3 + bx^2 + cx + d.

We'll identify the value of the leading coefficient as a = 1 and the value of b = -3.

The sum of the roots will yields:

x1 + x2 + x3 = -(-3)/1

x1 + x2 + x3 = 3

We could solve this problem, finding the roots of the polynomial first, then calculating their sum.

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

We'll factorize by x:

x(x^2 - 3x + 2) = 0

We'll cancel each factor:

x1 = 0

Since the sum of the roots of quadratic within brackets is 3 and the product is 2, the roots will be x2 = 1 and x3 = 2.

The usm will be x1 + x2 + x3 = 0 + 1 + 2 = 3.

**Therefore, the sum of the roots of the given polynomial is x1 + x2 + x3 = 3.**