Write the conjugate of `(-2-(1)/(3)i)^(3)`.

Answer

To find the conjugate of the complex number \((-2 - \frac{1}{3}i)^3\), we will follow these steps:

### Step 1: Rewrite the complex number
We start with the complex number:
\[ z = -2 - \frac{1}{3}i \]

### Step 2: Factor out \(-1\)
We can factor out \(-1\) from the expression:
\[ z = -1 \left(2 + \frac{1}{3}i\right) \]

### Step 3: Cube the expression
Now we will cube the expression:
\[ z^3 = (-1)^3 \left(2 + \frac{1}{3}i\right)^3 = -1 \left(2 + \frac{1}{3}i\right)^3 \]

### Step 4: Use the binomial expansion
Using the binomial expansion formula \((a + b)^3 = a^3 + 3a^2b + 3ab^2 + b^3\), where \(a = 2\) and \(b = \frac{1}{3}i\):
\[ \left(2 + \frac{1}{3}i\right)^3 = 2^3 + 3 \cdot 2^2 \cdot \frac{1}{3}i + 3 \cdot 2 \cdot \left(\frac{1}{3}i\right)^2 + \left(\frac{1}{3}i\right)^3 \]

### Step 5: Calculate each term
Calculating each term:
- \(2^3 = 8\)
- \(3 \cdot 2^2 \cdot \frac{1}{3}i = 3 \cdot 4 \cdot \frac{1}{3}i = 4i\)
- \(3 \cdot 2 \cdot \left(\frac{1}{3}i\right)^2 = 3 \cdot 2 \cdot \frac{-1}{9} = -\frac{6}{9} = -\frac{2}{3}\)
- \(\left(\frac{1}{3}i\right)^3 = \frac{1}{27}i^3 = -\frac{1}{27}i\)

### Step 6: Combine the terms
Now, combine all the terms:
\[ \left(2 + \frac{1}{3}i\right)^3 = 8 + 4i - \frac{2}{3} - \frac{1}{27}i \]
To combine the real parts:
\[ 8 - \frac{2}{3} = \frac{24}{3} - \frac{2}{3} = \frac{22}{3} \]
And for the imaginary parts:
\[ 4i - \frac{1}{27}i = \left(4 - \frac{1}{27}\right)i = \left(\frac{108}{27} - \frac{1}{27}\right)i = \frac{107}{27}i \]

### Step 7: Substitute back
So we have:
\[ \left(2 + \frac{1}{3}i\right)^3 = \frac{22}{3} + \frac{107}{27}i \]
Thus,
\[ z^3 = -\left(\frac{22}{3} + \frac{107}{27}i\right) = -\frac{22}{3} - \frac{107}{27}i \]

### Step 8: Find the conjugate
The conjugate of a complex number \(a + bi\) is \(a - bi\). Therefore, the conjugate of \(z^3\) is:
\[ \text{Conjugate} = -\frac{22}{3} + \frac{107}{27}i \]

### Final Answer
Thus, the conjugate of \((-2 - \frac{1}{3}i)^3\) is:
\[ -\frac{22}{3} + \frac{107}{27}i \]

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