Find the cube of: `3a - (1)/(a) (a ne 0)`

Answer

To find the cube of \( 3a - \frac{1}{a} \), we can use the formula for the cube of a binomial, which is given by:

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

In our case, we can identify:
- \( x = 3a \)
- \( y = -\frac{1}{a} \)

Now, we will apply the formula step by step.

### Step 1: Calculate \( x^3 \) and \( y^3 \)

1. Calculate \( x^3 = (3a)^3 \):
\[ (3a)^3 = 27a^3 \]

2. Calculate \( y^3 = \left(-\frac{1}{a}\right)^3 \):
\[ \left(-\frac{1}{a}\right)^3 = -\frac{1}{a^3} \]

### Step 2: Calculate \( 3x^2y \)

1. Calculate \( x^2 = (3a)^2 \):
\[ (3a)^2 = 9a^2 \]

2. Now calculate \( 3x^2y = 3 \cdot 9a^2 \cdot \left(-\frac{1}{a}\right) \):
\[ 3 \cdot 9a^2 \cdot \left(-\frac{1}{a}\right) = -27a \]

### Step 3: Calculate \( 3xy^2 \)

1. Calculate \( y^2 = \left(-\frac{1}{a}\right)^2 \):
\[ \left(-\frac{1}{a}\right)^2 = \frac{1}{a^2} \]

2. Now calculate \( 3xy^2 = 3 \cdot (3a) \cdot \left(\frac{1}{a^2}\right) \):
\[ 3 \cdot 3a \cdot \left(\frac{1}{a^2}\right) = \frac{9}{a} \]

### Step 4: Combine all parts

Now we can combine all the parts we calculated:
\[ (3a - \frac{1}{a})^3 = x^3 + y^3 + 3x^2y + 3xy^2 \]
Substituting the values we found:
\[ = 27a^3 - \frac{1}{a^3} - 27a + \frac{9}{a} \]

### Final Result

Thus, the final expression for the cube of \( 3a - \frac{1}{a} \) is:
\[ 27a^3 - \frac{1}{a^3} - 27a + \frac{9}{a} \]

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