If `3x+(1)/(2x)=5` then what is the value of `27x^(3)+(1)/(8x^(3))`?

Options

A. `102(1)/(2)`
B. `30(10)/(27)`
C. 0
D. 1

Answer

To solve the equation \(3x + \frac{1}{2x} = 5\) and find the value of \(27x^3 + \frac{1}{8x^3}\), we can follow these steps:

### Step 1: Cube both sides of the equation
We start with the equation:
\[ 3x + \frac{1}{2x} = 5 \]
Now, we cube both sides:
\[ \left(3x + \frac{1}{2x}\right)^3 = 5^3 \]

### Step 2: Apply the binomial expansion
Using the formula \((A + B)^3 = A^3 + B^3 + 3AB(A + B)\), where \(A = 3x\) and \(B = \frac{1}{2x}\):
\[ (3x)^3 + \left(\frac{1}{2x}\right)^3 + 3(3x)\left(\frac{1}{2x}\right)(3x + \frac{1}{2x}) = 125 \]

### Step 3: Calculate \(A^3\) and \(B^3\)
Calculating \(A^3\) and \(B^3\):
\[ (3x)^3 = 27x^3 \]
\[ \left(\frac{1}{2x}\right)^3 = \frac{1}{8x^3} \]
So, we have:
\[ 27x^3 + \frac{1}{8x^3} + 3(3x)\left(\frac{1}{2x}\right)(5) = 125 \]

### Step 4: Simplify the middle term
Now, simplify \(3(3x)\left(\frac{1}{2x}\right)(5)\):
\[ 3(3x)\left(\frac{1}{2x}\right) = \frac{9}{2} \]
Thus:
\[ 27x^3 + \frac{1}{8x^3} + \frac{9}{2} \cdot 5 = 125 \]
\[ 27x^3 + \frac{1}{8x^3} + \frac{45}{2} = 125 \]

### Step 5: Isolate \(27x^3 + \frac{1}{8x^3}\)
Now, subtract \(\frac{45}{2}\) from both sides:
\[ 27x^3 + \frac{1}{8x^3} = 125 - \frac{45}{2} \]

### Step 6: Convert \(125\) to a fraction
Convert \(125\) to have a common denominator of \(2\):
\[ 125 = \frac{250}{2} \]
So:
\[ 27x^3 + \frac{1}{8x^3} = \frac{250}{2} - \frac{45}{2} = \frac{205}{2} \]

### Step 7: Final answer
Thus, we have:
\[ 27x^3 + \frac{1}{8x^3} = 102.5 \]