Find all xx such that [x6x6 = (x+1)6(x+1)6].

Question

So far, I have found the real solution x= -1212, and the complex solution x = -−1‾‾‾√3−13.

Are there more, and if so what/how would be the most efficient find all the solutions to this problem? I am struggling to find the rest of the solutions.

Answer 1

The question latex was messed up, it is: Find all $x$ such that
\[x^6 = (x+1)^6.\]

Answer 2

according to your correction we simply have

x^6 = (x+1)^6

take 6th root of both sides, since we are taking an even root,
x = ±(x+1)
x = x+1 or x = -x-1
x = x+1 ---> no solution
or
x = -x-1
2x = -1
x = -1/2

check:
LS = (-1/2)^6 = 1/64
RS = (-1/2+ 1)^6
= (1/2)^6 = 1/64

x = -1/2

Answer 3

Yeah, that is the real solution. I also need to find the complex solution as well.

Answer 4

For the complex solutions, note that you want

x^6 - (x+1)^6 = 0

This is just the difference of two squares, so it factors as

(x^3 - (x+1)^3)(x^3 + (x+1)^3) = 0
The left factor becomes just a simple quadratic when expanded, as the x^3 terms go away. The right factor is the sum of two cubes, so it i

(x + (x+1))(x^2-x(x+1)+x(x+1))
The first part gives the real root, and the second becomes just another quadratic.

work it out, and you should come up with the complex roots shown here

http://www.wolframalpha.com/input/?i=x%5E6+%3D+(x%2B1)%5E6

Answer 5

oops. the cubes factor as

(x + (x+1))(x^2-x(x+1)+(x+1)^2)