I'm a Canadian Electrician - so CES is Canadian Electrical Code, but I'm sure physics in the US will be close.
Table 45 states a 3/4 HP motor draws 13.8A @ 115V or 6.9A @ 230V
Since "nominal" voltages are 120V and 240V - the motor is already assuming a 4% voltage drop.
From some informal on-line searches, people are suggesting the 3/4 HP pump will draw 6A - 8A. I supposes it depends how much load the water is putting on the pump (head & flow).
I'll use 10A @ 115V for this example (1,150W electric).
One characteristics of most motors is they continue to produce the same output even when provided with low voltage (unlike a resistive load/heater or incandescent lightbulb which will produce less output/light when voltage is less).
So, if voltage drop in your wire causes the voltage at the pump to drop to 103.5V (10% less than 115A rating - which is 13.75% voltage drop from the 120V supply) the amperage will increase by about the same percentage to maintain same power. So about 11A which is over-loading the pump windings by 10% - not great.
From CEC table D3 assuming the pump was fed by a #12 AWG wire a 400' run would cause the 13.75% voltage drop when the pump is drawing 11A.
So using the same 400' of wire, but now wired 240V (so the pump wants 230V @ 6.9A). Again using table D3 suggests 4% voltage drop. (note the actual voltage drop is about 9.6V, which would be 8% of 120V, but since we've upped the voltage, it is only 4% of 240V)
That 4% voltage drop - as mentioned earlier brings the 240V supply down to the motor rating of 230V at the bottom of the well, so you're both saving almost 10% in line losses (that saving 100W of power) you are now also running your pump in-spec, so less likely to over-heat or have early failure.
I hope that all makes sense!?! AND I hope I don't have any errors or typos in my calculations.