What is the vlaue of x satisfying the equation `16((a-x)/(a+x))^(3) = (a+x)/(a-x) ?`

Options

A. `a//2`
B. `a//3`
C. `a//4`
D. 0

Explanation

To solve the equation \( 16\left(\frac{a-x}{a+x}\right)^{3} = \frac{a+x}{a-x} \), we will follow these steps:

### Step 1: Rewrite the equation
We start with the given equation:
\[ 16\left(\frac{a-x}{a+x}\right)^{3} = \frac{a+x}{a-x} \]

### Step 2: Cross-multiply
To eliminate the fraction on the right side, we can cross-multiply:
\[ 16\left(\frac{a-x}{a+x}\right)^{3} \cdot (a-x) = (a+x) \]

### Step 3: Substitute \( y = \frac{a-x}{a+x} \)
Let \( y = \frac{a-x}{a+x} \). Then, we can rewrite the equation as:
\[ 16y^{3} = \frac{1}{y} \]

### Step 4: Multiply both sides by \( y \)
Multiplying both sides by \( y \) gives us:
\[ 16y^{4} = 1 \]

### Step 5: Solve for \( y \)
Now, we can solve for \( y \):
\[ y^{4} = \frac{1}{16} \]
Taking the fourth root of both sides:
\[ y = \frac{1}{2} \quad \text{or} \quad y = -\frac{1}{2} \]

### Step 6: Substitute back for \( x \)
Recall that \( y = \frac{a-x}{a+x} \). We will now solve for \( x \) in both cases.

#### Case 1: \( y = \frac{1}{2} \)
\[ \frac{a-x}{a+x} = \frac{1}{2} \]
Cross-multiplying gives:
\[ 2(a-x) = a+x \]
Expanding and rearranging:
\[ 2a - 2x = a + x \implies 2a - a = 2x + x \implies a = 3x \implies x = \frac{a}{3} \]

#### Case 2: \( y = -\frac{1}{2} \)
\[ \frac{a-x}{a+x} = -\frac{1}{2} \]
Cross-multiplying gives:
\[ -2(a-x) = a+x \]
Expanding and rearranging:
\[ -2a + 2x = a + x \implies 2x - x = a + 2a \implies x = 3a \implies x = -\frac{a}{3} \]

### Final Solution
Thus, the values of \( x \) satisfying the equation are:
\[ x = \frac{a}{3} \quad \text{and} \quad x = -\frac{a}{3} \]