Here's how.

The $\Delta V$ using instantaneous (e.g. chemical propulsion) maneuvers can be determined by repeated application of this equation that simply says that the total energy is the sum of the kinetic energy and the potential energy:

$\mathcal{E}=\frac{v^2}{2}-\frac{\mu}{r}$

where $\mathcal{E}$ is the total energy per unit mass of the object or the "specific energy", $v$ is the velocity of the object at the current position, $\mu$ is the GM of the central body, i.e. Newton's gravitational constant times its mass, and $r$ is the current distance from the center of the central body.

The key is that the total energy of the object is a constant of motion over the orbit.

We will also use the fact that orbits are ellipses, and this equation, which determines that constant of motion from the apses of the orbit, i.e. the radii of the closest and farthest points in the orbit, $r_1$ and $r_2$:

$\mathcal{E}=-{\mu\over r_1+r_2}$

For this problem we define:

$\mu_E$ = GM of the Earth.
$\mu_M$ = GM of the Moon.
$r_E$ = low Earth orbit radius.
$r_M$ = low Mars orbit radius.
$a_M$ = semi-major axis (mean radius) of Moon orbit around Earth.

For simplicity, we will assume that the Moon's orbit is circular.  You can reduce the $\Delta V$ a little by arranging to arrive at the Moon at its periapsis.

Using the above, in low-Earth orbit we have for the orbital velocity $v_{LEO}$:

$-{\mu_E\over 2r_E}=\frac{v_E^2}{2}-\frac{\mu_E}{r_E}$

which gives:

$v_{LEO}=\sqrt{\mu_E\over r_E}$

Similarly for the low-lunar orbit velocity $v_{LLO}$ we get:

$v_{LLO}=\sqrt{\mu_M\over r_M}$

And for the velocity of the Moon in its orbit around the Earth:

$v_M=\sqrt{\mu_E\over a_M}$

The Hohmann transfer from the Earth to the Moon is half of an elliptical orbit about the Earth with periapsis $r_M$ and apoapsis $a_M$.  For the velocity $v$ at any radius $r$ in that orbit, we have:

$-{\mu_E\over r_E+a_M}=\frac{v^2}{2}-\frac{\mu_E}{r}$

The velocity in that transfer orbit at the radius of a low-Earth orbit, i.e. its periapsis, is:

$v_p=\sqrt{2a_M \mu_E\over r_E(a_M+r_E)}$

Similarly, the velocity in that transfer orbit at the radius of the Moon, i.e. its apoapsis, is:

$v_a=\sqrt{2r_E \mu_E\over a_M(a_M+r_E)}$

The velocity to leave low-Earth orbit to get on the transfer orbit is:

$\Delta V_{inject}=v_p-v_{LEO}$

The velocity relative to the Moon on approach, if the Moon weren't there, is:

$v_\infty=v_M-v_a$

That is the velocity at infinity relative to the Moon, so there the specific energy relative to the Moon is:

$\mathcal{E}={v_\infty^2\over 2}$

(in progress -- pulled away for a bit, will finish later)