algebra precalculus - To show that 2 roots of $(x-a)^3$ + $(x-b)^3$ +$(x-c)^3 = 0 $ are complex

I am afraid that this question might be marked duplicate, but I simply had to ask if there is a way to show that the equation $(x-a)^3$ + $(x-b)^3$ +$(x-c)^3 = 0$ has only 1 real root? One of my teachers in high school posed the problem, and here's the twist, asked for a solution that does NOT invoke Rolle's theorem. He gave a hint that we could use the graph of said equation, but I don't see how that helps. I tried using Descartes' rules and Vieta's formulae, but they don't seem to lead anywhere. Thanks for helping!

EDIT:
After some insights from gracious responders,I see that my question boils down to this:Can we construct some artfully watertight piece of argument which shows that an increasing function has one and only one real root?