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

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:

$a^3 - b^3 = (a-b)(a^2 + ab + b^2)$

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

The equation becomes:

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

Multiply each term by $(m-3)(m^2+3m+9)$ to eliminate the denominators, assuming $m \neq 3$:

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

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

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

Combine like terms with respect to $x$:

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

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

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

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

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

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

$(m+3)x = -1$

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

If $m = 3$, then the equation becomes:

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

$(9-9)x + 0 = 0$

$0x = 0$

This means that when $m=3$, any value of $x$ is a solution, however $m=3$ makes the denominators in the original equation equal to zero so the solution $m=3$ and any value of $x$ is not valid.

Also, $m$ cannot be equal to $-3$, otherwise, $x = \frac{-1}{m+3}$ will be undefined.

3. Final Answer

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