The tedious way is to expand $(x_1 - x_2)^2 (x_2 - x_3)^2 (x_1 - x_3)^2$ out completely and then write it in terms of $x_1 + x_2 + x_3$ and so forth. You are guaranteed that this is possible by the fundamental theorem of symmetric polynomials, the proof of which even gives an algorithm for doing this, but it's a pain to do by hand (although it's not a bad exercise in algebraic manipulation).
A less tedious way is to argue as follows. We will first work under the assumption that $p = 0$. Now, $q, r$ are polynomials of degrees $2, 3$, and the discriminant is a polynomial of degree $6$, so the discriminant must be a linear combination of the two monomials $q^3, r^2$. Thus we can write
$$\Delta = a q^3 + b r^2$$
for two constants $a, b$, where $\Delta$ is the discriminant. Setting $q = -1, r = 0$ we get the polynomial $x^3 - x = 0$ with roots $0, 1, -1$. We compute that the discriminant is equal to $4$, from which it follows that $a = -4$.
Setting $q = 0, r = -1$ we get the polynomial $x^3 - 1 = 0$ with roots $1, \omega, \omega^2$ where $\omega$ is a primitive third root of unity. Using the identity
$$(\omega - 1)^2 = \omega^2 - 2 \omega + 1 = - 3 \omega$$
we compute that the discriminant is equal to $-27$, from which it follows that $b = -27$. Thus
$$\Delta = -4 q^3 - 27 r^2.$$
To get from here to an arbitrary choice of $x_1, x_2, x_3$, apply the above formula to the polynomial with roots $x_1 - \frac{p}{3}, x_2 - \frac{p}{3}, x_3 - \frac{p}{3}$ and note that subtracting the same constant from each of the three roots doesn't change the discriminant.
Consistent with the comment of dxiv,
let $~y = x-1,~$ and let $~u = y^2.$
Then
$$(y^2 - 4)(y^2 - 36) = 1680 \implies $$
$$u^2 - 40u - 1536 = 0 \implies \tag1 $$
$$u = \frac{1}{2} ~\left[40 \pm \sqrt{1600 + (4 \times 1536)}\right] \tag2 $$
$$ = \frac{1}{2} \left[40 \pm 88\right] = 20 \pm 44.$$
Since you are looking for real solutions, you must have that
$$y^2 = u = 64 \implies (x-1) = y = \pm 8.$$
Addendum
Actually, all of the Math, following (2) above is unnecessary. From (2) above, it is immediate that the value inside the radical will be bigger than $~(40)^2,~$ and so only one (positive) value of $~y^2 = u~$ will be forthcoming.
Then regardless of the eventual computed value of $~u = y^2,~$ you will have the two real roots of
$$x-1 = y = \pm \sqrt{u}.$$
So, denoting the two roots as $~x_1,~$ and $~x_2,~$ you have that $~\displaystyle (x_1 - 1) + (x_2 - 1) = \sqrt{u} + (-\sqrt{u}) = 0.$
So, without bothering to solve for $~u,~$ you know that $~x_1 + x_2 = 2.$
In fact, with experience, you can see that because the third term in (1) above, $~-1536,~$ is negative, you could (instead) stop the Math after (1) above, rather than (2) above.
That is, (1) above, with any negative 3rd term, is sufficient to imply that only one positive real root of $~y^2 = u~$ will be forthcoming.
Best Answer
Before using Vieta’s theorem, the equation $$x^3-2x^2+2x-1=\frac{15}{x}$$ must first be rewritten as a polynomial: $$\implies x^4-2x^3+2x^2-x-15=0$$ (notice that $x\neq 0$). The sum of its roots is indeed $-\frac{b}{a}=-\frac{-2}{1}=2$.
Even so, the question asked for the sum of its real roots, whereas Vieta’s formulae give the sum of its complex (including real) roots. That is, the above “sum of its roots” might have contained something like $1+i$ and $1-i,$ and two other roots that cancel each other.
Thus, instead of Vieta's theorem, this exercise requires another technique, such as factoring, as you said, or by using the rational roots theorem.