#1**+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

CPhill May 20, 2019

#2**+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.**

Guest May 20, 2019