PLZ HELP ME!!!
Find all integers n for which n^3 = (n-1)^3+(n-2)^3+(n-3)^3.
THANK YOU!
+23245 Multiply out the cubes:
(n - 1)3 = n3 - 3n2 + 3n - 1
(n - 2)3 = n3 - 6n2 + 12n - 8
(n - 3)3 = n3 - 9n2 + 27n - 27
Substitute them into the original problem:
n3 = (n - 1)3 + (n - 2)3 + (n - 3)3
n3 = [ n3 - 3n2 + 3n - 1 ] + [ n3 - 6n2 + 12n - 8 ] + [ n3 - 9n2 + 27n - 27 ]
Simplify: 0 = 2n3 - 18n2 + 42n - 36 ---> 0 = n3 - 9n2 + 21n - 18
This can be factored; however, if you cannot find the factor -- and, if the solution is an integer, you
can try each of the factors of -18 -- 1, -1, 2, -2, 3, -3, 6, -6, 9, -9, 18, -18
One ot these is the correct answer ...
