Olympiad Mathematics

Hi,

[abrahamgaladima]

It appears Mr. Jake that 8 is not a solution to the problem. I tried to do a check but it didn’t work out. Thanks Sir.

[QqQq-ec]

I suggest to substitute variable x by: y=x-8
NB: the equation becomes even

We get:
(y+8) + (y-8) + 3 [y(y+8) (y-8)]^1/3 = y
=> y^3=-27 y(y^2 - 64)
Case 1- solution 1: y= 0
---
Case 2: 28 y^2 = 27 * 64
Solution 2: y=+12 (3/7)^1/2 ;
Solution 3: y=-12 (3/7)^1/2

[satrajitghosh]

writing u = x^ (1/3)
v = ( x - 16)^(1/3)
w = ( x - 8)^(1/3)
we get
u^3 + v^3 = 2 w^3
u + v = w
u ^3 + v^3 + 3 u v ( u + v)
= 2 u ^ 3 + 2 v^3
(u + v) ( u ^2 - 4 u v + v^2) = 0
u = - v, (2 + √3) v, (2 - √3) v
Case I ( u = v)
x ^ (1/3) = (x - 16) ^ (1/3)
No feasible solution
Case II ( u = (2 + √3) v)
x ^ (1/3) * (2 - √3) = (x - 16) ^ (1/3)
x * (2 + √3)^3 = ( x - 16)
x ( 8 - 3 √ 3 - 12 √3 + 18) = x - 16
x (15 √3 - 25) = 16
x = 16 * (15 √3 + 25) / (725 - 625)
= 4 * ( 3 √ 3 + 5)/5
= 12 √ 3/5 + 4
Case II ( u = (2 - √3) v)
x = 4 - 12 √ 3/5