Which is larger


(2^1/3-1)^1/3     or     (1/9)^1/3 - (2/9)^1/3 + (4/9)^1/3 ?

Cube both numbers.  The first is just  2^(1/3)-1
The second is
1/9 - 2/9 + 4/9
- 3*(1/9)^(2/3)*(2/9)^(1/3)
+ 3*(1/9)^(2/3)*(4/9)^(1/3)
+ 3*(2/9)^(2/3)*(1/9)^(1/3)
+ 3*(2/9)^(2/3)*(4/9)^(1/3)
+ 3*(4/9)^(2/3)*(1/9)^(1/3)
- 3*(4/9)^(2/3)*(2/9)^(2/3)
- 6*- 3*(1/9)^(1/3)*(2/9)^(1/3)*(4/9)^(1/3)
= simplify
1/9 - 2/9 + 4/9
- 2^(1/3)/3 + 2^(2/3)/3 + 2^(2/3)/3
+ 2^(4/3)/3 + 2^(4/3)/3 - 2^(5/3)/3
- 2^(2)/3
= combine like terms and add numerators
[1 - 2^(1/3) + 2^(7/3) - 2^(2))]/3
= factor
[1 - 2^(1/3) - 2^(2)*(1 - 2^(1/3))]/3
= factor
[(1-2^(2))*(1 - 2^(1/3))]/3
= simplify
[(-3)*(1 - 2^(1/3))]/3
= simplify
2^(1/3) - 1
which is the same as the cube of the first number.  The answer the question:  They are equal.

Each of the original numbers is just an expression for the real cube root of 2^(1/3) - 1

Posted by Jer on 2013-05-21 14:11:23