If `(f(x))^2*f((1-x)/(1+x))=64 x AA in D_f` then

Options

A. `4x^(2//3)((1+x)/(1-x))^(1//3)`
B. `x^(1//3)((1-x)/(1+x))^(1//3)`
C. `x^(1//3)((1-x)/(1+x))^(1//3)`
D. `x((1+x)/(1-x))^(1//3)`

Explanation

To solve the equation \((f(x))^2 \cdot f\left(\frac{1-x}{1+x}\right) = 64x\), we will follow these steps:

### Step 1: Write the original equation
We start with the given equation:
\[ (f(x))^2 \cdot f\left(\frac{1-x}{1+x}\right) = 64x \]
Let’s denote this as Equation (1).

### Step 2: Substitute \(y = \frac{1-x}{1+x}\)
We will make a substitution to simplify our calculations. Let:
\[ y = \frac{1-x}{1+x} \]
From this, we can express \(x\) in terms of \(y\):
\[ x = \frac{1-y}{1+y} \]
Now, we can rewrite the original equation using this substitution.

### Step 3: Rewrite the equation
Substituting \(y\) into Equation (1), we get:
\[ (f(y))^2 \cdot f\left(\frac{1-y}{1+y}\right) = 64 \cdot \frac{1-y}{1+y} \]
This can be rewritten as:
\[ (f(y))^2 \cdot f(y) = 64 \cdot \frac{1-y}{1+y} \]
Let’s denote this as Equation (2).

### Step 4: Square both sides of Equation (1)
Now, we will square Equation (1):
\[ ((f(x))^2)^2 \cdot \left(f\left(\frac{1-x}{1+x}\right)\right)^2 = (64x)^2 \]
This simplifies to:
\[ (f(x))^4 \cdot f\left(\frac{1-x}{1+x}\right)^2 = 4096x^2 \]

### Step 5: Divide by Equation (2)
Now, we will divide the squared version of Equation (1) by Equation (2):
\[ \frac{(f(x))^4 \cdot f\left(\frac{1-x}{1+x}\right)^2}{(f(y))^2 \cdot f\left(\frac{1-y}{1+y}\right)} = \frac{4096x^2}{64 \cdot \frac{1-y}{1+y}} \]
This simplifies to:
\[ \frac{(f(x))^4}{(f(y))^2} = 64 \cdot \frac{1+y}{1-y} \cdot x^2 \]

### Step 6: Solve for \(f(x)\)
From the previous steps, we can express \(f(x)\):
\[ f(x)^3 = 64x^2 \cdot \frac{1+y}{1-y} \]
This gives us:
\[ f(x) = 4x^{2/3} \cdot \left(\frac{1+y}{1-y}\right)^{1/3} \]

### Conclusion
Thus, the function \(f(x)\) can be expressed as:
\[ f(x) = 4x^{2/3} \cdot \left(\frac{1+\frac{1-x}{1+x}}{1-\frac{1-x}{1+x}}\right)^{1/3} \]