I stumbled across this identity purely by accident:

sqrt(MC - XI) = sqrt(IX) + sqrt(CM)This is rather elegant in Roman numerals! Is it merley a coincidence, or is something deeper going on here?

For example, does the equation hold in other bases?
For this question to make sense, the values I, X, C, and M should
be interpreted as 1, b, b^{2} and b^{3},
where b is the base.
The proposed identity is then:

sqrt((bThis certainly doesn't look to be generally true, but in fact, it is!^{3}+ b^{2}) - (b + 1)) = sqrt(b - 1) + sqrt(b^{3}- b^{2})

LHS = sqrt(bSurprisingly, the identity holds for all bases. Of course, each expression is integral only when b is one more than a square.^{2}* (b + 1) - (b + 1)) = sqrt((b + 1) * (b^{2}- 1)) = sqrt((b + 1) * (b + 1) * (b - 1)) = (b + 1) * sqrt(b - 1) = b * sqrt(b - 1) + sqrt(b - 1) = sqrt(b^{3}- b^{2}) + sqrt(b - 1) = RHS