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I am interested in finding a general method of solving equations involving cube roots such as $$x^{1/3} + (x-16)^{1/3} = (x-8)^{1/3}.$$

I have a solution for this particular one:

$$\{8 - (12 \cdot 21^{1/2})/7, \quad 8 + (12 \cdot 21^{1/2})/7, \quad 8\}$$

but it only worked because of the $0,8,16$ connection to the x's which made a final simplification possible.

I'm sure that someone in the past must have done some work on this sort of equation. I know of Cardano's work and wonder if there is a connection?

Thanks for any help.

TMM
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user132599
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    It is probably pretty hopeless. Change your $8$ to a $7$. With natural manipulations we end up with a polynomial equation of degree $\gt 4$, with no symmetries to bring down the degree. – André Nicolas Mar 02 '14 at 22:18
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    I believe such efforts ultimately founder on the fact that $ \ (a + b)^n \ \neq \ a^n + b^n \ , \ n > 1 . $ We readily solve equations like $ \ \sqrt{a} + \sqrt{b} \ = \ \sqrt{c} \ $ ($ \ a, b, c \ $ being expressions) because you can square both sides, and the binomial-square creates only one radical that can then be isolated and "squared-away" on a second pass. The problem with doing this on your equation is that "cubing both sides" is going to give you two cube-roots on one side, with correspondingly more work to eliminate the radicals. This only gets worse for still higher roots. – colormegone Mar 02 '14 at 22:48

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