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

 

Please put your second question on a different post - Melody.

 May 23, 2019
edited by Melody  May 24, 2019
 #1
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1.

 

n^3  = ( n - 1)^3 +( n - 2)^3 + ( n - 3)^3

 

Simplify

 

n^3  =  3n^3 -18n^2 + 42n - 36     rearrange as

 

2n^3 - 18n^2 + 42n - 36  = 0        divide throught by 2

 

n^3 - 9n^2 + 21n - 18  =  0    [ split up the 9n^2 term as 6n^2 + 3n^2 ]

 

n^3 - 6n^2 + 3n^2 +21n - 18  = 0     this factors as

 

n^2 ( n - 6) + 3(n^2 + 7n - 6)  = 0

 

n^2(n - 6)  + 3 (n - 6) ( n + 1) = 0

 

(n - 6 ) [ n^2 + 3(n + 1) )  = 0

 

(n - 6) [ n^2 + 3n + 3]  = 0

 

The second factor has no real solutions when set to 0  [ the discriminant is negative ]

 

So

 

n - 6  =  0

 

n = 6

 

 

 

cool cool cool 

 May 24, 2019
edited by CPhill  May 24, 2019

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