If `x gt 0 and x^(2) +(1)/(9x^(2))= (25)/(36)`, find `x^(3) + (1)/(27x^(3))`

Answer

To solve the equation \( x^2 + \frac{1}{9x^2} = \frac{25}{36} \) and find \( x^3 + \frac{1}{27x^3} \), we can follow these steps:

### Step 1: Rewrite the expression
We start with the equation:
\[ x^2 + \frac{1}{9x^2} = \frac{25}{36} \]
We notice that \( x^2 + \frac{1}{9x^2} \) can be related to \( \left( x + \frac{1}{3x} \right)^2 \).

### Step 2: Express \( x^2 + \frac{1}{9x^2} \) in terms of \( \left( x + \frac{1}{3x} \right)^2 \)
Using the identity:
\[ \left( x + \frac{1}{3x} \right)^2 = x^2 + 2 \cdot x \cdot \frac{1}{3x} + \left( \frac{1}{3x} \right)^2 = x^2 + \frac{2}{3} + \frac{1}{9x^2} \]
We can rearrange this to find:
\[ x^2 + \frac{1}{9x^2} = \left( x + \frac{1}{3x} \right)^2 - \frac{2}{3} \]

### Step 3: Substitute back into the equation
Substituting this back into our equation gives:
\[ \left( x + \frac{1}{3x} \right)^2 - \frac{2}{3} = \frac{25}{36} \]
Adding \( \frac{2}{3} \) to both sides:
\[ \left( x + \frac{1}{3x} \right)^2 = \frac{25}{36} + \frac{2}{3} \]
To add these fractions, we convert \( \frac{2}{3} \) to have a common denominator of 36:
\[ \frac{2}{3} = \frac{24}{36} \]
Thus:
\[ \left( x + \frac{1}{3x} \right)^2 = \frac{25 + 24}{36} = \frac{49}{36} \]

### Step 4: Take the square root
Taking the square root of both sides:
\[ x + \frac{1}{3x} = \frac{7}{6} \]
Since \( x > 0 \), we only consider the positive root.

### Step 5: Cube both sides
Now we cube both sides:
\[ \left( x + \frac{1}{3x} \right)^3 = \left( \frac{7}{6} \right)^3 \]
Using the expansion formula:
\[ a^3 + b^3 + 3ab(a + b) \]
where \( a = x \) and \( b = \frac{1}{3x} \), we have:
\[ x^3 + \frac{1}{27x^3} + 3 \cdot x \cdot \frac{1}{3x} \left( x + \frac{1}{3x} \right) = \frac{343}{216} \]
This simplifies to:
\[ x^3 + \frac{1}{27x^3} + \left( x + \frac{1}{3x} \right) = \frac{343}{216} \]

### Step 6: Substitute \( x + \frac{1}{3x} \)
We know from earlier that \( x + \frac{1}{3x} = \frac{7}{6} \). Substitute this in:
\[ x^3 + \frac{1}{27x^3} + \frac{7}{6} = \frac{343}{216} \]

### Step 7: Isolate \( x^3 + \frac{1}{27x^3} \)
Now, we need to isolate \( x^3 + \frac{1}{27x^3} \):
\[ x^3 + \frac{1}{27x^3} = \frac{343}{216} - \frac{7}{6} \]
Convert \( \frac{7}{6} \) to a fraction with a denominator of 216:
\[ \frac{7}{6} = \frac{252}{216} \]
Thus:
\[ x^3 + \frac{1}{27x^3} = \frac{343 - 252}{216} = \frac{91}{216} \]

### Final Answer
Therefore, the value of \( x^3 + \frac{1}{27x^3} \) is:
\[ \boxed{\frac{91}{216}} \]