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The implicit expression $(b-a)=(a+b)^3$ looks like it could be written explicitly for $a$ as a function of $b$. The only region of interest is for $a,b>0$ Here is what the plot looks like:

alt text

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    $\begingroup$ You cannot express $b$ as an explicit function of $a$, even on the first quadrant, because $b$ is not a function of $a$: the portion of the graph does not pass the vertical line test. On the other hand $a$ does seem like it would be a function of $b$. $\endgroup$ Nov 8, 2010 at 20:48
  • $\begingroup$ Well, first you have to note that the vertical axis crosses the graph of your equation thrice... $\endgroup$ Nov 8, 2010 at 20:48
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    $\begingroup$ You have a cubic in b. There are explicit (but messy) formulas for the solution of a cubic. Try Wikipedia. $\endgroup$ Nov 8, 2010 at 20:50
  • $\begingroup$ I did mean $a$ as a function of $b$. I have edited my original post. $\endgroup$
    – Gus
    Nov 8, 2010 at 20:50
  • $\begingroup$ Ok, you now have a cubic with a single real root (for $a$). Again, you can use the explicit solution for the cubic. $\endgroup$ Nov 8, 2010 at 20:52

1 Answer 1

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Writing $p=a+b$ you have the cubic equation, $$ p^3+p-2b=0. $$ This is already in "depressed cubic" form (no $p^2$ term), so it can be solved directly by standard methods. The coefficient of p is positive, so it is strictly increasing and there will be a single real root. $$ p = \sqrt[3]{\sqrt{b^2+1/27}+b}-\sqrt[3]{\sqrt{b^2+1/27}-b} $$ or, $$ a = \sqrt[3]{\sqrt{b^2+1/27}+b}-\sqrt[3]{\sqrt{b^2+1/27}-b}-b. $$ Alternatively, using the hyperbolic method, $$ a=\frac{2}{\sqrt{3}}\sinh\left(\frac13\sinh^{-1}\left(3\sqrt{3}b\right)\right)-b. $$

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  • $\begingroup$ +1: For including the hyperbolic method. I will delete my answer. $\endgroup$
    – Aryabhata
    Nov 8, 2010 at 21:24
  • $\begingroup$ @George Lowther thanks for your answer! $\endgroup$
    – Gus
    Nov 8, 2010 at 21:28
  • $\begingroup$ @Moron Did you delete your answer because it was incorrect or because it was more complicated? I thought it was interesting... $\endgroup$
    – Gus
    Nov 8, 2010 at 21:30
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    $\begingroup$ @Moron: I think I have to side with Gus here. I don't think there's any need to delete a good answer just because someone else has posted something similar. $\endgroup$ Nov 8, 2010 at 21:33
  • $\begingroup$ @Gus: I deleted it because it is a subset of George's answer. If you look at the link to hyperbolic method, it has what I had written (I wasn't aware of that when I wrote my answer). @George: Your answer is a superset, not just similar :-) If it had something different, I would have kept it around. $\endgroup$
    – Aryabhata
    Nov 8, 2010 at 21:33

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