A nice little problem.
Assume it is possible; then (2^(1/3)  1)^(1/3) = (b/k)^(1/3)+(c/k)^(1/3)+(d/k)^(1/3), with b,c,d,k integers.
The formula isn't telling us much as it is, so say k^(1/3) = 2^(1/3), then LHS, when multiplied, is 1/3(2+2^(1/3)2^(2/3)) 3^(1/3), with 3 terms, which looks significant in context.
Splitting up the LHS [1/3*2*3^(1/3)]+[1/3*2^(1/3)*3^(1/3)][1/3*2^(2/3))*3^(1/3)].
So we have:
b^(1/3)2^(1/3)=1/3*2*3^(1/3), cubing everything 2b = 8/9, b=4/9,
c^(1/3)2^(1/3)=1/3*2^(1/3)*3^(1/3), 2c=2/9, c=1/9,
d^(1/3)2^(1/3)=1/3*2^(2/3)) 3^(1/3), 2d=4/9, d=2/9.
And indeed (2^(1/3)  1)^(1/3) = (1/9)^(1/3)(2/9)^(1/3)+(4/9)^(1/3). It's always nice to find a rational expression for cube roots!
Edited on October 24, 2015, 10:32 pm

Posted by broll
on 20151024 11:29:46 