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

 May 20, 2019
 #1
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Expanding the right side....we have....

 

 

3 n^3 - 18 n^2 + 42 n - 36  = n^3

 

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

 

n^3 - 9n^2 + 21n - 18  = 0

 

There is only one integer root at x = 6

 

See the graph, here : https://www.desmos.com/calculator/90pyxjmftb

 

 

cool cool cool

 May 20, 2019
 #2
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Solve for n:
n^3 = (n - 3)^3 + (n - 2)^3 + (n - 1)^3

Expand out terms of the right hand side:
n^3 = 3 n^3 - 18 n^2 + 42 n - 36

Subtract 3 n^3 - 18 n^2 + 42 n - 36 from both sides:
-2 n^3 + 18 n^2 - 42 n + 36 = 0

The left hand side factors into a product with three terms:
-2 (n - 6) (n^2 - 3 n + 3) = 0

Divide both sides by -2:
(n - 6) (n^2 - 3 n + 3) = 0

Split into two equations:
n - 6 = 0 or n^2 - 3 n + 3 = 0
n = 6. The quaratic equation has complex roots only.

 May 20, 2019

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