The problem is to solve the equation $\frac{2mx(m+1)+3x}{m^3-27} - \frac{x}{m-3} + \frac{x+1}{m^2+3m+9} = 0$ for $x$.

1. Problem Description

The problem is to solve the equation

$\frac{2mx(m+1)+3x}{m^3-27} - \frac{x}{m-3} + \frac{x+1}{m^2+3m+9} = 0$ for $x$.

2. Solution Steps

First, we factor the denominator $m^3 - 27$ using the difference of cubes formula:

$m^3 - 27 = m^3 - 3^3 = (m-3)(m^2 + 3m + 9)$.

Now, rewrite the given equation with the factored denominator:

$\frac{2mx(m+1)+3x}{(m-3)(m^2+3m+9)} - \frac{x}{m-3} + \frac{x+1}{m^2+3m+9} = 0$

Next, we find a common denominator for all the terms, which is $(m-3)(m^2+3m+9)$:

$\frac{2mx(m+1)+3x}{(m-3)(m^2+3m+9)} - \frac{x(m^2+3m+9)}{(m-3)(m^2+3m+9)} + \frac{(x+1)(m-3)}{(m-3)(m^2+3m+9)} = 0$

Combine the numerators over the common denominator:

$\frac{2mx(m+1)+3x - x(m^2+3m+9) + (x+1)(m-3)}{(m-3)(m^2+3m+9)} = 0$

For the fraction to be zero, the numerator must be zero:

$2mx(m+1)+3x - x(m^2+3m+9) + (x+1)(m-3) = 0$

Expand the terms:

$2m^2x + 2mx + 3x - xm^2 - 3mx - 9x + xm - 3x + m - 3 = 0$

Combine like terms:

$(2m^2 - m^2)x + (2m - 3m + m)x + (3 - 9 - 3)x + m - 3 = 0$

$m^2x + 0x - 9x + m - 3 = 0$

$m^2x - 9x + m - 3 = 0$

Factor out $x$:

$(m^2 - 9)x = 3 - m$

Factor the difference of squares:

$(m-3)(m+3)x = 3 - m$

$(m-3)(m+3)x = -(m-3)$

Now, we can solve for $x$:

If $m \neq 3$, divide both sides by $(m-3)(m+3)$:

$x = \frac{-(m-3)}{(m-3)(m+3)}$

$x = \frac{-1}{m+3}$

The condition $m \neq 3$ is necessary, otherwise the original expression has undefined terms.

Also, we need $m \neq -3$, otherwise we are dividing by zero.

3. Final Answer

$x = -\frac{1}{m+3}$ for $m \neq 3$ and $m \neq -3$.