Express the following in the form a+bi
`(-2-(1)/(3)i)^(3)`

Answer

To express the complex number \((-2 - \frac{1}{3}i)^{3}\) in the form \(a + bi\), we will follow these steps:

### Step 1: Rewrite the expression
We start with the expression:
\[ (-2 - \frac{1}{3}i)^{3} \]
We can rewrite this as:
\[ (-1)^{3} \cdot (2 + \frac{1}{3}i)^{3} = - (2 + \frac{1}{3}i)^{3} \]

### Step 2: Use the binomial expansion
Using the binomial expansion for \((A + B)^{3}\), where \(A = 2\) and \(B = \frac{1}{3}i\):
\[ (A + B)^{3} = A^{3} + B^{3} + 3AB(A + B) \]
Substituting \(A\) and \(B\):
\[ (2 + \frac{1}{3}i)^{3} = 2^{3} + \left(\frac{1}{3}i\right)^{3} + 3 \cdot 2 \cdot \frac{1}{3}i \cdot (2 + \frac{1}{3}i) \]

### Step 3: Calculate each term
Calculating \(2^{3}\):
\[ 2^{3} = 8 \]
Calculating \(\left(\frac{1}{3}i\right)^{3}\):
\[ \left(\frac{1}{3}i\right)^{3} = \frac{1}{27}i^{3} = \frac{1}{27}(-i) = -\frac{1}{27}i \]
Calculating \(3 \cdot 2 \cdot \frac{1}{3}i \cdot (2 + \frac{1}{3}i)\):
\[ 3 \cdot 2 \cdot \frac{1}{3}i = 2i \]
Now, we need to calculate \(2i(2 + \frac{1}{3}i)\):
\[ 2i \cdot 2 + 2i \cdot \frac{1}{3}i = 4i + \frac{2}{3}i^{2} \]
Since \(i^{2} = -1\):
\[ \frac{2}{3}i^{2} = -\frac{2}{3} \]
So, we have:
\[ 2i(2 + \frac{1}{3}i) = 4i - \frac{2}{3} \]

### Step 4: Combine all terms
Now we combine all the terms:
\[ (2 + \frac{1}{3}i)^{3} = 8 - \frac{1}{27}i + 4i - \frac{2}{3} \]
Combining the real parts:
\[ 8 - \frac{2}{3} = \frac{24}{3} - \frac{2}{3} = \frac{22}{3} \]
Combining the imaginary parts:
\[ -\frac{1}{27}i + 4i = -\frac{1}{27}i + \frac{108}{27}i = \frac{107}{27}i \]

### Step 5: Final expression
Thus, we have:
\[ (2 + \frac{1}{3}i)^{3} = \frac{22}{3} + \frac{107}{27}i \]
So,
\[ - (2 + \frac{1}{3}i)^{3} = -\frac{22}{3} - \frac{107}{27}i \]

### Final Answer
The expression in the form \(a + bi\) is:
\[ -\frac{22}{3} - \frac{107}{27}i \]