A contractor has 40 men to do a job in 60 days working 9 hours a day. How many more men are to be employed to finish the job in 48 days working 10 hours a day?

The total work is proportional to the number of men, days, and hours per day.
Let $M_1$ be the initial number of men, $D_1$ be the initial number of days, and $H_1$ be the initial hours per day.
$M_1 = 40$ men
$D_1 = 60$ days
$H_1 = 9$ hours/day

Total work units $= M_1 \times D_1 \times H_1$

$$\text{Total work} = 40 \times 60 \times 9 = 2400 \times 9 = 21600 \text{ work units}$$

Let $M_2$ be the new number of men, $D_2$ be the new number of days, and $H_2$ be the new hours per day.
$D_2 = 48$ days
$H_2 = 10$ hours/day

Since the total work remains the same:

$$M_2 \times D_2 \times H_2 = \text{Total work}$$ $$M_2 \times 48 \times 10 = 21600$$ $$M_2 \times 480 = 21600$$

$$M_2 = \frac{21600}{480}$$ $$M_2 = \frac{2160}{48}$$ $$M_2 = 45 \text{ men}$$

This is the total number of men needed to complete the job under the new conditions.

The contractor initially employed 40 men. The new requirement is 45 men.
Number of more men needed $= M_2 - M_1$

$$\text{More men} = 45 - 40 = 5 \text{ men}$$

The contractor needs to employ 5 more men.

---

The given equation is:

$$\frac{3^3}{9} = \frac{1}{3 - 2x}$$

Calculate $3^3$:

$$3^3 = 3 \times 3 \times 3 = 27$$

Substitute this value back into the left side:

$$\frac{27}{9} = 3$$

Multiply both sides by $(3 - 2x)$:

$$3(3 - 2x) = 1$$

Distribute the 3 on the left side:

$$9 - 6x = 1$$

Subtract 9 from both sides:

$$-6x = 1 - 9$$ $$-6x = -8$$

Divide both sides by -6:

$$x = \frac{-8}{-6}$$

Simplify the fraction:

$$x = \frac{8}{6}$$ $$x = \frac{4}{3}$$

The final answer for the word problem is $\boxed{\text{5 men}}$.
The final answer for the equation is $\boxed{x = \frac{4}{3}}$.