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
[(12^(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 20130521 14:11:23 