Hey Dennis,
you have trhee-phase electricity on the output of the spindle, each wave shifted by 120° at the same time. Also the 1.5 kW spindle that the VFD is for is by no means the electrical power draw but the mechanical power that is delivered at the shaft. The electrical power draw is much higher. I have shown in an example how to compute the power draw of a spindlle in in this posting for a VFD for 2.2 kW spindle. If you follow the computation example, what you have in the end is the electrical current draw at the spindle input = the output of the VFD, which is much higher than 1.5 kVA. THEN, for the current draw at the input of the VFD you still have to add the effficiency loss of the VFD itself. A VFD is an AC-to-DC-to-three-phase-AC inverter. It recitifies all AC at its input and stores it in a large capacitor, and then with six IGBTs, it switches the three phases on and off in a quasi-sine waves on three phases on three wires, with a variable frequency. If it has a single phase input, all the current comes in one wire, while at the output, the power is spread over three phases simultanously, so the current rating of the spindle (what is printed on the spindle), i.e. the current in one wire is usually a lower number. See also this posting here.
Also the formula for three-phase current is not P = V · I, but P = V · √3 · I.