If there is only $1$ complex solution to the equation $8x^3 + 12x^2 + kx + 1 = 0$, what is $k$?
1. Given the cubic equation \(8x^3 + 12x^2 + kx + 1 = 0\), we need to find the value of \(k\) such that there is exactly one complex solution.
2. For a cubic equation to have exactly one complex solution, it must have a triple root. This means the equation can be written in the form \(8(x - \alpha)^3 = 0\) for some complex number \(\alpha\).
3. Expanding \(8(x - \alpha)^3\), we get:
   \[
   8(x - \alpha)^3 = 8(x^3 - 3\alpha x^2 + 3\alpha^2 x - \alpha^3)
   \]
   Simplifying, we have:
   \[
   8(x - \alpha)^3 = 8x^3 - 24\alpha x^2 + 24\alpha^2 x - 8\alpha^3
   \]
4. Comparing this with the given equation \(8x^3 + 12x^2 + kx + 1 = 0\), we equate the coefficients:
   \[
   8x^3 - 24\alpha x^2 + 24\alpha^2 x - 8\alpha^3 = 8x^3 + 12x^2 + kx + 1
   \]
5. From the coefficient of \(x^2\), we get:
   \[
   -24\alpha = 12 \implies \alpha = -\frac{1}{2}
   \]
6. Substituting \(\alpha = -\frac{1}{2}\) into the coefficient of \(x\), we get:
   \[
   24\alpha^2 = k \implies 24\left(-\frac{1}{2}\right)^2 = k \implies 24 \cdot \frac{1}{4} = k \implies k = 6
   \]
7. Finally, we check the constant term:
   \[
   -8\alpha^3 = 1 \implies -8\left(-\frac{1}{2}\right)^3 = 1 \implies -8 \cdot -\frac{1}{8} = 1 \implies 1 = 1
   \]
   This confirms that our value of \(\alpha\) and \(k\) are correct.

The final answer is \(\boxed{6}\).