If `3x- (4)/(x)= 4 and x ne 0`, find `27 x^(3)- (64)/(x^(3))`

Answer

To solve the equation \(3x - \frac{4}{x} = 4\) and find \(27x^3 - \frac{64}{x^3}\), we will follow these steps:

### Step 1: Rearranging the Given Equation
Start with the equation:
\[ 3x - \frac{4}{x} = 4 \]
Rearranging gives:
\[ 3x = 4 + \frac{4}{x} \]

### Step 2: Square Both Sides
Square both sides of the equation to eliminate the fraction:
\[ (3x - \frac{4}{x})^2 = 4^2 \]
Using the identity \((a - b)^2 = a^2 + b^2 - 2ab\):
\[ (3x)^2 + \left(\frac{4}{x}\right)^2 - 2(3x)\left(\frac{4}{x}\right) = 16 \]
This simplifies to:
\[ 9x^2 + \frac{16}{x^2} - 24 = 16 \]

### Step 3: Simplifying the Equation
Now, rearranging gives:
\[ 9x^2 + \frac{16}{x^2} = 16 + 24 \]
\[ 9x^2 + \frac{16}{x^2} = 40 \]

### Step 4: Finding \(27x^3 - \frac{64}{x^3}\)
We need to express \(27x^3 - \frac{64}{x^3}\) using the identity:
\[ a^3 - b^3 = (a - b)(a^2 + ab + b^2) \]
Let \(a = 3x\) and \(b = \frac{4}{x}\). Then:
\[ 27x^3 - \frac{64}{x^3} = (3x - \frac{4}{x})\left((3x)^2 + (3x)(\frac{4}{x}) + \left(\frac{4}{x}\right)^2\right) \]
Substituting \(3x - \frac{4}{x} = 4\):
\[ = 4\left((3x)^2 + 3x \cdot \frac{4}{x} + \left(\frac{4}{x}\right)^2\right) \]

### Step 5: Calculating Each Term
Calculating each term:
- \((3x)^2 = 9x^2\)
- \(3x \cdot \frac{4}{x} = 12\)
- \(\left(\frac{4}{x}\right)^2 = \frac{16}{x^2}\)

Thus:
\[ = 4\left(9x^2 + 12 + \frac{16}{x^2}\right) \]

### Step 6: Substitute \(9x^2 + \frac{16}{x^2}\)
From Step 3, we know:
\[ 9x^2 + \frac{16}{x^2} = 40 \]
So:
\[ = 4(40 + 12) = 4 \cdot 52 \]

### Step 7: Final Calculation
Calculating:
\[ 4 \cdot 52 = 208 \]

### Final Answer
Thus, the value of \(27x^3 - \frac{64}{x^3}\) is:
\[ \boxed{208} \]