1. **State the problem:** Find the points of intersection of the curves given by the equations: $$y = x^2 - 3x + x$$ $$y = x^3 - 3x^2 + x$$ 2. **Simplify the equations:** For the first curve: $$y = x^2 - 3x + x = x^2 - 2x$$ So the system is: $$y = x^2 - 2x$$ $$y = x^3 - 3x^2 + x$$ 3. **Set the two expressions for $y$ equal to find intersection points:** $$x^2 - 2x = x^3 - 3x^2 + x$$ 4. **Bring all terms to one side:** $$0 = x^3 - 3x^2 + x - x^2 + 2x$$ $$0 = x^3 - 4x^2 + 3x$$ 5. **Factor the polynomial:** $$0 = x(x^2 - 4x + 3)$$ 6. **Factor the quadratic:** $$x^2 - 4x + 3 = (x - 3)(x - 1)$$ 7. **Find roots:** $$x = 0, x = 1, x = 3$$ 8. **Find corresponding $y$ values using $y = x^2 - 2x$:** - For $x=0$: $$y = 0^2 - 2(0) = 0$$ - For $x=1$: $$y = 1^2 - 2(1) = 1 - 2 = -1$$ - For $x=3$: $$y = 3^2 - 2(3) = 9 - 6 = 3$$ 9. **Final answer:** The points of intersection are: $$\boxed{(0,0), (1,-1), (3,3)}$$