Find the cube of: `2a + (1)/(2a) (a ne 0)`

Answer

To find the cube of \(2a + \frac{1}{2a}\), we will use the formula for the cube of a binomial, which is given by:

\[ (x + y)^3 = x^3 + y^3 + 3x^2y + 3xy^2 \]

Here, we can identify \(x = 2a\) and \(y = \frac{1}{2a}\).

### Step 1: Identify \(x\) and \(y\)
Let:
- \(x = 2a\)
- \(y = \frac{1}{2a}\)

### Step 2: Calculate \(x^3\) and \(y^3\)
Now we calculate \(x^3\) and \(y^3\):
\[ x^3 = (2a)^3 = 8a^3 \]
\[ y^3 = \left(\frac{1}{2a}\right)^3 = \frac{1}{8a^3} \]

### Step 3: Calculate \(3x^2y\)
Next, we calculate \(3x^2y\):
\[ 3x^2y = 3(2a)^2\left(\frac{1}{2a}\right) = 3(4a^2)\left(\frac{1}{2a}\right) = 3 \cdot 2a = 6a \]

### Step 4: Calculate \(3xy^2\)
Now, we calculate \(3xy^2\):
\[ 3xy^2 = 3(2a)\left(\frac{1}{2a}\right)^2 = 3(2a)\left(\frac{1}{4a^2}\right) = \frac{3 \cdot 2}{4} = \frac{3}{2} \]

### Step 5: Combine all parts
Now we can combine all the parts together:
\[ (2a + \frac{1}{2a})^3 = x^3 + y^3 + 3x^2y + 3xy^2 \]
Substituting the values we calculated:
\[ = 8a^3 + \frac{1}{8a^3} + 6a + \frac{3}{2} \]

### Step 6: Final expression
Thus, the final expression for the cube of \(2a + \frac{1}{2a}\) is:
\[ (2a + \frac{1}{2a})^3 = 8a^3 + 6a + \frac{1}{8a^3} + \frac{3}{2} \]