Answer
Stable equilibrium at $x=\frac{1}{6}$.
Work Step by Step
We can determine the required equilibrium positions along with stability at these positions as follows:
As given that, $V=8x^3-2x^2-10$
At equilibrium, we have
$\frac{dV}{dx}=0$
$\implies \frac{d(8x^3-2x^2-10)}{dx}=24x^2-4x=0$ eq(1)
This simplifies to:
$x_1=0$, $x_2=\frac{1}{6}$
We take the second derivative of eq(1)
$\frac{d^2V}{dx^2}=48x-4$
for $x=\frac{1}{6}$, the above equation becomes
$\frac{d^2V}{dx^2}=48(\frac{1}{6})-4=4\gt 0$
which means stable equilibrium.
for $x=0$, we get:
$\frac{d^2V}{dx^2}=48(0)-4=-4\lt 0$, unstable
