Find the only positive integer whose cube is the sum of the cubes of three positive integers immediately preceding it. Find this positive integer. Your algebraic work must be detailed enough to show this is the only positive integer with this property

Answer 1

The property translates to the following equation:

x^3-(x-1)^3-(x-2)^3-(x-3)^3=0
Expanding:
-2x^3+18x^2-42x+36 = 0
2(-x³+9x²-24x+18)=0
which factorizes to:
-2(x-6)(x^2-3x+3)=0
or simply:
(x-6)(x^2-3x+3)=0

The first factor gives x=6 (our answer), and the second factor is not further factorizable.
Attempts to solve
(x^2-3x+3)=0
results in a complex number, so no other real roots exist.