We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive pseudonymised information about your use of our website. cookie policy and privacy policy.
 
+0  
 
+1
44
2
avatar

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

 May 20, 2019
 #1
avatar+101420 
+3

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
avatar
+1

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

12 Online Users

avatar
avatar
avatar