Find the stationary points of: $V=(2/r^3)-(3/r^2)$

Question

I've applied the exponent rules and differentiated to get to the point where i have

$$dr/dv=-6r^{-4}+6r^{-3}$$

I'm starting to get very confused with the math when i set the LHS = 0 in order to find the value of stationary points.

Any guidance would be helpful.

Answer 1

So we have $$-6r^{-4}+6r^{-3}=0\implies 6r^{-4}=6r^{-3}\implies {6\over r^4}={6\over r^3}\;\;/\cdot r^4$$

$$\implies 6=6r\implies r=1$$