
Ng Kwong Ming asked,
How can we proceed to obtain
(2^{1/3}1)^{1/3} = (1/9)^{1/3}  (2/9)^{1/3} + (4/9)^{1/3} ?
Cubing the RHS only verify the question and sheds no light on the solution of the problem.
Marcos noted that the RHS is a short geometric progression,
S = a+ab+ab^2, where
a=(1/9)^{1/3}
b=(2)^{1/3}
Simplifying it,
S = a+ab+ab^{2}
bS = ab+ab^{2}+ab^{3}
bSS=ab^{3}a
S(b1)=a(b^{3}1)
S=a(b^{3}1)/(b1)
S=(1/9)^{1/3} (3)/(2^{1/3}1)
so it is equal to
3^{1/3}/(1+2^{1/3})
Now we have (2^{1/3}1)^{1/3} = (1/9)^{1/3}  (2/9)^{1/3} + (4/9)^{1/3} = 3^{1/3}/(1+2^{1/3}).
Multiplying the cube of the left side by 3 times the cube of the reciprocal of the right side,
(2^{1/3} + 1)^{3}(2^{1/3}  1)
= (2^{2/3}  1)(2^{2/3} + (2)2^{1/3} + 1)
= (2^{2/3})(2^{2/3} + (2)2^{1/3} + 1)  (2^{2/3} + (2)2^{1/3} + 1)
= (2)2^{1/3} + 4 + 2^{2/3}  2^{2/3}  (2)2^{1/3}  1
= 3
Another approach would be to cube the new RHS, 3^{1/3}/(1+2^{1/3}). I'll start with the denominator:
(1+2^{1/3})^{3}
= 1 + (3)2^{1/3} + (3)2^{2/3} + 2
= (3)2^{1/3} + (3)2^{2/3} + (3). . . .
Factoring Cubic, Continued  It's a curious fact that expressions involving nested square and cube roots, which come up in connection with the cubic formula, can be simplified by factoring the cubic equation in question. For example,
_________
³√10 + Ö 108 __________
³Ö 10 + Ö 108can be simplified to an integer: 2. This page does a thorough analysis of expressions of this type that simplify to integers.
Recurrence Relation of Successive Powers of Polynomial Root  what a mouthful! The gist of this topic is that the successive powers of a root of a polynomial form a sequence that has an simple recurrence relation.
The webmaster and author of this Math Help site is Graeme McRae.