The problem is to solve the inequality $\frac{x^2 - 3x + 1}{x-3} > 0$.

1. Problem Description

The problem is to solve the inequality $\frac{x^2 - 3x + 1}{x-3} > 0$.

2. Solution Steps

First, we need to find the zeros of the numerator, $x^2 - 3x + 1 = 0$. We can use the quadratic formula to find the roots:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

In our case, $a = 1$, $b = -3$, and $c = 1$. Thus,

$x = \frac{3 \pm \sqrt{(-3)^2 - 4(1)(1)}}{2(1)}$

$x = \frac{3 \pm \sqrt{9 - 4}}{2}$

$x = \frac{3 \pm \sqrt{5}}{2}$

So, the roots of the numerator are $x_1 = \frac{3 - \sqrt{5}}{2} \approx 0.38$ and $x_2 = \frac{3 + \sqrt{5}}{2} \approx 2.62$.

The denominator is $x-3$. The value that makes the denominator zero is $x = 3$.

Now, we consider the intervals defined by these critical points: $(-\infty, \frac{3 - \sqrt{5}}{2})$, $(\frac{3 - \sqrt{5}}{2}, \frac{3 + \sqrt{5}}{2})$, $(\frac{3 + \sqrt{5}}{2}, 3)$, and $(3, \infty)$. We will test a point in each interval to determine the sign of the expression $\frac{x^2 - 3x + 1}{x-3}$.

1. Interval $(-\infty, \frac{3 - \sqrt{5}}{2})$: Let $x = 0$.

$\frac{0^2 - 3(0) + 1}{0 - 3} = \frac{1}{-3} = -\frac{1}{3} < 0$

2. Interval $(\frac{3 - \sqrt{5}}{2}, \frac{3 + \sqrt{5}}{2})$: Let $x = 1$.

$\frac{1^2 - 3(1) + 1}{1 - 3} = \frac{1 - 3 + 1}{-2} = \frac{-1}{-2} = \frac{1}{2} > 0$

3. Interval $(\frac{3 + \sqrt{5}}{2}, 3)$: Let $x = 2.7$.

$\frac{(2.7)^2 - 3(2.7) + 1}{2.7 - 3} = \frac{7.29 - 8.1 + 1}{-0.3} = \frac{0.19}{-0.3} < 0$

4. Interval $(3, \infty)$: Let $x = 4$.

$\frac{4^2 - 3(4) + 1}{4 - 3} = \frac{16 - 12 + 1}{1} = \frac{5}{1} = 5 > 0$

The inequality $\frac{x^2 - 3x + 1}{x-3} > 0$ is satisfied in the intervals $(\frac{3 - \sqrt{5}}{2}, \frac{3 + \sqrt{5}}{2})$ and $(3, \infty)$.

3. Final Answer

The solution to the inequality is $(\frac{3 - \sqrt{5}}{2}, \frac{3 + \sqrt{5}}{2}) \cup (3, \infty)$.