No....we can't do that.....
We have
(x + 1)^16 / x^16 = 1 multiply bothsides by x^16
(x + 1)^16 = x^16 subtract x^16 from both sides
(x + 1)^16 - x^16 = 0 we can factor this several times, as follows
[ ( x + 1)^8 + x^8 ] [ (x + 1)^8 - x ^8] = 0
The first factor won't have a real solution.....so.....working with the second
[ (x + 1)^4 + x^4] [ (x + 1)^4 - x^4] = 0
Again, the first has no real solution.....factor the second
[ ( x+ 1)^2 + x^2] [ (x + 1)^2 - x^2] = 0
No real solution for the first.......factor the second one more time
[ ( x + 1) + x] [ (x + 1) - x ] = 0
The second has no real solution......set the first to 0
(x + 1) + x = 0 subtract 1 from both sides
2x = -1 divide both sides by 2
x = -1/2 this is the only real solution
