The image contains a mathematical expression:

$\frac{3x^3}{x^2 - 2x - 3} + \frac{5x}{x^2 - 2x - 3} = \frac{2x + 6}{x^2 - 2x - 3}$

To solve this equation, we can start by combining the fractions on the left-hand side:

$\frac{3x^3 + 5x}{x^2 - 2x - 3} = \frac{2x + 6}{x^2 - 2x - 3}$

Since the denominators are the same, we can equate the numerators:

$3x^3 + 5x = 2x + 6$

Rearranging terms to bring all terms to one side gives:

$3x^3 + 5x - 2x - 6 = 0$

$3x^3 + 3x - 6 = 0$

We can factor out a 3 from the entire equation:

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

$x^3 + x - 2 = 0$

Now, we need to solve the cubic equation $x^3 + x - 2 = 0$. This can be done by trying possible rational roots or using more advanced methods. Let's test $x = 1$ as a potential solution:

$1^3 + 1 - 2 = 1 + 1 - 2 = 0$

Thus, $x = 1$ is a root. To find the other roots, we can factor $x - 1$ out of $x^3 + x - 2$:

Using synthetic division or polynomial long division to divide $x^3 + x - 2$ by $x - 1$, we get:

$(x - 1)(x^2 + x + 2) = 0$

Now, solve $x^2 + x + 2 = 0$:

Using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:

$x = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 1 \cdot 2}}{2 \cdot 1} = \frac{-1 \pm \sqrt{1 - 8}}{2} = \frac{-1 \pm \sqrt{-7}}{2}$

$x = \frac{-1 \pm \sqrt{7}i}{2}$

The roots are $x = 1$ and $x = \frac{-1 \pm \sqrt{7}i}{2}$.

Would you like to see more details, or do you have any specific questions?

Here are some related questions to consider:

1. How do you perform polynomial long division to divide two polynomials?
2. What are the possible methods for solving cubic equations?
3. How can synthetic division help in finding roots of polynomials?
4. What is the significance of imaginary roots in polynomials?
5. How can the Rational Root Theorem help identify potential roots of a polynomial?

Tip: Always verify potential rational roots by substituting them back into the original polynomial equation.