We are asked to solve the equation $\frac{3}{x^2 - 64} = \frac{4}{4x^2 + 32x}$ for $x$.

1. Problem Description

We are asked to solve the equation $\frac{3}{x^2 - 64} = \frac{4}{4x^2 + 32x}$ for $x$.

2. Solution Steps

First, we factor the denominators:

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

$4x^2 + 32x = 4x(x+8)$

Thus, the equation can be written as

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

We multiply both sides by $4x(x-8)(x+8)$ to clear the denominators:

$3 \cdot 4x = 4(x-8)$

$12x = 4x - 32$

$12x - 4x = -32$

$8x = -32$

$x = \frac{-32}{8}$

$x = -4$

Now we check if $x=-4$ is a valid solution. The original equation is

$\frac{3}{x^2 - 64} = \frac{4}{4x^2 + 32x}$.

When $x=-4$,

$x^2 - 64 = (-4)^2 - 64 = 16 - 64 = -48$

$4x^2 + 32x = 4(-4)^2 + 32(-4) = 4(16) - 128 = 64 - 128 = -64$

The equation becomes

$\frac{3}{-48} = \frac{4}{-64}$

$\frac{3}{-48} = -\frac{1}{16}$

$\frac{4}{-64} = -\frac{1}{16}$

So, the equation holds true.

3. Final Answer

x = -4