Find all integers for which \( $$n^3 = (n-1)^3+(n-2)^3+(n-3)^3.$$\)
n^3 = (n-1)^3+(n-2)^3+(n-3)^3.
I could only find one value for n:
n = 6
6^3 = 5^3 + 4^3 + 3^3 = 216
Yes, but there are 2 more that you can't find by bashing it.
n^3 = (n-1)^3+(n-2)^3+(n-3)^3
\(n^3 = (n-1)^3+(n-2)^3+(n-3)^3\\ n^3 = (n^3-3n^2+3n-1)+(n^3-3*2n^2+3*4n-8)+(n^3-3*3n^2+3*9n-27)\\ n^3 = n^3-3n^2+3n-1+n^3-6n^2+12n-8+n^3-9n^2+27n-27\\ 0= 2n^3 -18n^2 +42n -36 \\ 0= n^3 -9n^2 +21n -18 \\\)
You can take if from there.
I got three solutions:
1. \(6\)
2. \(\frac {3+i\sqrt3}2\)
3. \(\dfrac {3-1\sqrt3}2\)
You can also write it as
\(6, \dfrac {3\pm i\sqrt3}2\)
+23 
