The water bill (R) is partly constant and partly varies as the cube of the volume (V). If R=4.25 when V=50 and R=5.16 when V=60, find the expression for R in terms of V.

The problem states that the water bill $R$ is partly constant and partly varies as the cube of the volume $V$. This can be written as:

$$R = C + kV^3$$

where $C$ is the constant part and $k$ is the constant of proportionality.

Scenario 1: When $V = 50$ litres, $R = 4.25$ cedis.
Substitute these values into the equation:

$$4.25 = C + k(50)^3$$ $$4.25 = C + 125000k \quad (1)$$

Scenario 2: When $V = 60$ litres, $R = 5.16$ cedis.
Substitute these values into the equation:

$$5.16 = C + k(60)^3$$ $$5.16 = C + 216000k \quad (2)$$

Subtract equation (1) from equation (2):

$$(C + 216000k) - (C + 125000k) = 5.16 - 4.25$$ $$91000k = 0.91$$

Divide by 91000 to find $k$:

$$k = \frac{0.91}{91000}$$ $$k = 0.00001$$

Now substitute the value of $k$ into equation (1) to find $C$:

$$4.25 = C + 125000(0.00001)$$ $$4.25 = C + 1.25$$

Subtract 1.25 from both sides to find $C$:

$$C = 4.25 - 1.25$$ $$C = 3.00$$

Substitute the values of $C$ and $k$ back into the general equation $R = C + kV^3$:

$$R = 3 + 0.00001V^3$$

The expression for $R$ in terms of $V$ is $\boxed{R = 3 + 0.00001V^3}$.