If the six roots of `x^6 = -64` are written in the form `a + ib`, where a and b are real, then the product of those roots for which `a lt 0` is

Explanation

To solve the problem of finding the product of the roots of the equation \(x^6 = -64\) for which \(a < 0\), we will follow these steps:

### Step 1: Rewrite the equation
We start with the equation:
\[ x^6 = -64 \]
We can rewrite \(-64\) as:
\[ -64 = 64 \cdot i^2 \]
Thus, we have:
\[ x^6 = 64 \cdot i^2 \]

### Step 2: Express in polar form
Next, we express \(64\) in polar form. The modulus is \(64\) and the argument is \(\pi\) (since \(-64\) is on the negative real axis). Therefore:
\[ x^6 = 64 \text{cis}(\pi) \]
where \(\text{cis}(\theta) = \cos(\theta) + i\sin(\theta)\).

### Step 3: Find the sixth roots
To find the sixth roots, we use the formula for roots in polar form:
\[ x_k = r^{1/n} \text{cis}\left(\frac{\theta + 2k\pi}{n}\right) \]
where \(k = 0, 1, 2, \ldots, n-1\). Here, \(r = 64\), \(\theta = \pi\), and \(n = 6\):
\[ x_k = 2 \text{cis}\left(\frac{\pi + 2k\pi}{6}\right) = 2 \text{cis}\left(\frac{\pi}{6} + \frac{k\pi}{3}\right) \]

### Step 4: Calculate the roots
Now we can calculate the six roots for \(k = 0, 1, 2, 3, 4, 5\):
- For \(k = 0\):
\[ x_0 = 2 \text{cis}\left(\frac{\pi}{6}\right) = 2\left(\cos\frac{\pi}{6} + i\sin\frac{\pi}{6}\right) = 2\left(\frac{\sqrt{3}}{2} + i\frac{1}{2}\right) = \sqrt{3} + i \]

- For \(k = 1\):
\[ x_1 = 2 \text{cis}\left(\frac{\pi}{2}\right) = 2i \]

- For \(k = 2\):
\[ x_2 = 2 \text{cis}\left(\frac{5\pi}{6}\right) = 2\left(-\frac{\sqrt{3}}{2} + i\frac{1}{2}\right) = -\sqrt{3} + i \]

- For \(k = 3\):
\[ x_3 = 2 \text{cis}\left(\frac{7\pi}{6}\right) = 2\left(-\frac{\sqrt{3}}{2} - i\frac{1}{2}\right) = -\sqrt{3} - i \]

- For \(k = 4\):
\[ x_4 = 2 \text{cis}\left(\frac{3\pi}{2}\right) = -2i \]

- For \(k = 5\):
\[ x_5 = 2 \text{cis}\left(\frac{11\pi}{6}\right) = 2\left(\frac{\sqrt{3}}{2} - i\frac{1}{2}\right) = \sqrt{3} - i \]

### Step 5: Identify roots with \(a < 0\)
The roots are:
1. \(\sqrt{3} + i\) (Real part \(> 0\))
2. \(2i\) (Real part \(= 0\))
3. \(-\sqrt{3} + i\) (Real part \(< 0\))
4. \(-\sqrt{3} - i\) (Real part \(< 0\))
5. \(-2i\) (Real part \(= 0\))
6. \(\sqrt{3} - i\) (Real part \(> 0\))

The roots for which \(a < 0\) are:
- \(-\sqrt{3} + i\)
- \(-\sqrt{3} - i\)

### Step 6: Calculate the product of these roots
The product of the roots is:
\[ (-\sqrt{3} + i)(-\sqrt{3} - i) = (-\sqrt{3})^2 - (i)^2 = 3 - (-1) = 3 + 1 = 4 \]

### Final Answer
Thus, the product of the roots for which \(a < 0\) is:
\[ \boxed{4} \]