The equation is $$\\ \sqrt[3]{\sqrt{a}+b}+\sqrt[3]{-\sqrt{a}+b}=k.$$
How do I find$\ a$?
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Had a chance to look at my answer? – Gerry Myerson Oct 13 '14 at 06:19
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1@GerryMyerson Uh, sorry, though it is true that yesterday I wasn't home, I checked my notifications but no sign of your answer. Thank you! – Vincenzo Oliva Oct 13 '14 at 12:16
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There are some nice cancellations. It will look less imposing if we let $c=\sqrt a$ Then we have $$\sqrt[3]{c+b}+\sqrt[3]{-c+b}=k\\ (c+b)+(b-c)+3\sqrt[3]{2b(b^2-c^2)}=k^3\\3\sqrt[3]{2b(b^2-c^2)}=k^3-2b$$ an you are on your way.
Ross Millikan
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How do you get$\ 3\sqrt[3]{2b\left(b^2-c^2\right)}$ from$\ 3\sqrt[3]{\left(b+c\right)^2(b-c)}+3\sqrt[3]{\left(b-c\right)^2(b+c)}$? – Vincenzo Oliva Oct 08 '14 at 05:49
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$$k=(b+c)^{1/3}+(b-c)^{1/3}$$ where $c=\sqrt a$. We use $(x+y)^3=x^3+y^3+3xy(x+y)$.
$$k^3=b+c+b-c+3(b^2-c^2)^{1/3}k=2b+3k(b^2-c^2)^{1/3}$$
Now subtract $2b$, divide by $3k$, cube, subtract from $b^2$, and you have $c^2$, which is $a$.
Gerry Myerson
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