+0  
 
+1
29
1
avatar

PLZ HELP ME!!!

Find all integers n for which n^3 = (n-1)^3+(n-2)^3+(n-3)^3.

 

THANK YOU!

 Aug 1, 2020
 #1
avatar+21943 
+1

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 ...

 
 Aug 1, 2020

11 Online Users

avatar
avatar
avatar