If `2x + (2)/(9x) = 4,` then the value of `27 x ^(3) + (1)/( 27 x ^(3))` is

Options

A. 180
B. 198
C. 234
D. 252

Answer

To solve the equation \(2x + \frac{2}{9x} = 4\) and find the value of \(27x^3 + \frac{1}{27x^3}\), we can follow these steps:

### Step 1: Simplify the given equation
Start with the equation:
\[ 2x + \frac{2}{9x} = 4 \]
To eliminate the fraction, multiply both sides by \(9x\):
\[ 9x(2x) + 9x\left(\frac{2}{9x}\right) = 9x(4) \]
This simplifies to:
\[ 18x^2 + 2 = 36x \]

### Step 2: Rearrange the equation
Rearranging gives:
\[ 18x^2 - 36x + 2 = 0 \]

### Step 3: Solve the quadratic equation
To solve \(18x^2 - 36x + 2 = 0\), we can use the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \(a = 18\), \(b = -36\), and \(c = 2\). Plugging in these values:
\[ x = \frac{36 \pm \sqrt{(-36)^2 - 4 \cdot 18 \cdot 2}}{2 \cdot 18} \]
Calculating the discriminant:
\[ (-36)^2 = 1296 \]
\[ 4 \cdot 18 \cdot 2 = 144 \]
So,
\[ x = \frac{36 \pm \sqrt{1296 - 144}}{36} \]
\[ x = \frac{36 \pm \sqrt{1152}}{36} \]
Calculating \(\sqrt{1152}\):
\[ \sqrt{1152} = 24\sqrt{2} \]
Thus,
\[ x = \frac{36 \pm 24\sqrt{2}}{36} = 1 \pm \frac{2\sqrt{2}}{3} \]

### Step 4: Find \(3x + \frac{1}{3x}\)
Let \(y = 3x\). Then:
\[ y = 3\left(1 \pm \frac{2\sqrt{2}}{3}\right) = 3 \pm 2\sqrt{2} \]
Now we calculate:
\[ 3x + \frac{1}{3x} = y + \frac{1}{y} \]
Using the identity:
\[ y + \frac{1}{y} = 6 \quad \text{(as derived from the earlier steps)} \]

### Step 5: Find \(27x^3 + \frac{1}{27x^3}\)
Using the identity:
\[ y^3 + \frac{1}{y^3} = (y + \frac{1}{y})^3 - 3(y + \frac{1}{y}) \]
Substituting \(y + \frac{1}{y} = 6\):
\[ y^3 + \frac{1}{y^3} = 6^3 - 3 \cdot 6 \]
Calculating:
\[ 6^3 = 216 \quad \text{and} \quad 3 \cdot 6 = 18 \]
Thus:
\[ y^3 + \frac{1}{y^3} = 216 - 18 = 198 \]

### Final Answer
The value of \(27x^3 + \frac{1}{27x^3}\) is:
\[ \boxed{198} \]