Express the expression: ³√(³√2 - 1) in the form:
³√x + ³√y + ³√z, where each of x, y and z is a rational number.

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 2015-10-24 11:29:46