which all goes to remind us that one size does not fit all, and that each individual case has to be designed on its own merits
Going a bit deeper into the maths
The heater rated 9.75 Kw at 230 has a resistance of 5.42 ohms
R = V*V/W = 230 * 230 / 9750 = 5.42
The cable ( based on quoted volt drop ) has a resistance of 0.27 ohms
R = V / I = 11.45 / 42 = 11.45 / 42 = 0.272
Total circuit resistance is therefore 5.69 ohms
With 230 volts applied the current will be 40.42 amps
I = V / R = 230 / 5.69 = 40.42
Power dissipated as heat in shower will be 8855 watts
W = R*I*I = 5.42 * 40.42 * 40.42 = 8855.068088
Power dissipated as heat in the cable will be 441.12 watts
W = R*I*I = 0.27 * 40.42 * 40.42 = 441.119
Total power = 8855 + 441 = 9296 watts
% useful in heater 95.25 %
% wasted in cable 4.74 %
Shower is running at 91 % of rated heat output
8.85 Kw and not at 9.75 Kw as expected.
Power loss in cable is significant when the designed cable size is just inside the voltage drop limits. For a shower the duration of the loss is fixed by the person in the shower, but for applications such as heating a kettle or immersion heated tank of water the duration depends on how much heat is reaching the water. Some one else can do the financial maths of what would be the pay back time for investing in cable larger than the minimum size the voltage drop regulations require to be used.
And the thought of 12 watts per metre being used to heat a copper wire inside two layers of PVC insulation (441 / 37 = 11.91 ) tends to worry me. The idea of that amount of heat depending on being clipped to a wall to help dissipate the heat leads me to use larger cable,