Solve the equation $\frac{4}{x^2-8x} - \frac{3}{x^2+8x} = \frac{x}{x^2-64}$.

1. Problem Description

Solve the equation $\frac{4}{x^2-8x} - \frac{3}{x^2+8x} = \frac{x}{x^2-64}$.

2. Solution Steps

First, factor the denominators of the given equation:

$x^2 - 8x = x(x-8)$

$x^2 + 8x = x(x+8)$

$x^2 - 64 = (x-8)(x+8)$

The equation becomes

$\frac{4}{x(x-8)} - \frac{3}{x(x+8)} = \frac{x}{(x-8)(x+8)}$

The common denominator for all terms is $x(x-8)(x+8)$. Multiply both sides of the equation by $x(x-8)(x+8)$. Note that $x$ cannot be $0$, $8$, or $-8$, as this will make the denominators zero.

$x(x-8)(x+8) \left(\frac{4}{x(x-8)} - \frac{3}{x(x+8)}\right) = x(x-8)(x+8) \left(\frac{x}{(x-8)(x+8)}\right)$

$4(x+8) - 3(x-8) = x^2$

$4x + 32 - 3x + 24 = x^2$

$x + 56 = x^2$

$x^2 - x - 56 = 0$

Now, we solve the quadratic equation $x^2 - x - 56 = 0$. We can factor the quadratic as follows:

$(x-8)(x+7) = 0$

So, $x=8$ or $x=-7$.

However, we noted earlier that $x$ cannot be $8$ or $-8$, since that would make the denominator zero. Thus, $x=8$ is an extraneous solution and must be excluded.

Therefore, the only solution is $x=-7$.

3. Final Answer

$x=-7$