The equation $x^3 + 8x^2 - 4x + c = 0$ has three roots, one of which is the product of the other two. What is c?
Let the roots be $a,$ $b,$ and $ab.$ Then by Vieta's formulas,
\[a + b + ab = -8,\]so $ab + a + b = -8.$ Completing the square in $a$ and $b,$ we get
\[(a + 1)(b + 1) = -7.\]The other root is then $-(a + b + 1) = ab.$ Hence,
\[c = -7 \cdot ab = \boxed{7}.\]
Let's call the three roots of the equation $r_1$, $r_2$, and $r_3$, with $r_1$ being the product of $r_2$ and $r_3$.
We know that the sum of the roots of a cubic equation can be found by taking the opposite sign of the coefficient of the quadratic term (in this case $-8$) and dividing it by the coefficient of the cubic term (in this case $1$). So, the sum of the roots is $-(-8)/1 = 8$.
Since $r_1 = r_2 \times r_3$, we can rewrite the equation as $x^3 + 8x^2 - 4x + r_2 \times r_3 = 0$.
The sum of the roots also equals the opposite sign of the coefficient of the $x^2$ term ($8$) divided by the coefficient of the $x$ term ($-4$). So, we have $8/(-4)=-2$.
From this, we can conclude that the sum of the roots $r_1$, $r_2$, and $r_3$ is $-2$. Since $r_1 = r_2 \times r_3$ and the only way to get a sum of $-2$ from three roots is by having two negative and one positive root, we can assume that the positive root is $r_1$, and the two negative roots are $r_2$ and $r_3$.
Therefore, $r_1 + r_2 + r_3 = -2$ becomes $r_1 - (r_1 + r_2) = -2$, which simplifies to $-r_2 = -2$. Solving for $r_2$, we have $r_2 = 2$.
We know that $r_1 = r_2 \times r_3$, so $2 = r_2 \times r_3$. Since $r_2 = 2$, we have $2 = 2 \times r_3$, which implies $r_3 = 1$.
Finally, substituting the values of $r_1$, $r_2$, and $r_3$ into the original equation $x^3 + 8x^2 - 4x + c = 0$, we have $x^3 + 8x^2 - 4x + c = (x - r_1)(x - r_2)(x - r_3)$. Plugging in the values, we get $(x - 2)(x - 1)(x - 1)$. Multiplying this out, we obtain $x^3 - 4x^2 + 5x - 2 = 0$.
Comparing this with the original equation, we see that $c = \boxed{-2}$.