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I want to solve the following equation for $x$ or $y$ (does not matter wich one) analytically.

$$\sqrt[3]{x+y} + \sqrt[3]{x-y} = 1$$

Wolframalpha returns following solution, but I could not think of a way how to get there:

$$ x+1 \neq 0 \qquad y = \frac{(x+1) \sqrt{8 x-1}}{3 \sqrt{3}}$$

Is there a nice 'tool' I do not know for solving this?

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up vote 3 down vote accepted

$$(\sqrt[3]{x+y} + \sqrt[3]{x-y})^3 = 1$$

$$x+y+3(\sqrt[3]{x+y}\cdot\sqrt[3]{x-y})\underbrace{(\sqrt[3]{x+y} + \sqrt[3]{x-y})}_1+x-y=1$$





$$27y^2 = 8 x^3+15 x^2+6 x-1$$

$$27y^2 = (x+1)^2 (8 x-1)$$

$$y = \frac{(x+1) \sqrt{8 x-1}}{3 \sqrt{3}}$$

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Thank you very much! The idea of factoring out the 1 is really neat, and was the key to where I failed. – flawr May 17 '14 at 21:55

Suppose $a+b+c=0$ and that $a,b,c$ are the roots of the equation $$f(x)=x^3-px^2+qx-r=0$$

Then $p=a+b+c=0, r=abc$ and adding $f(a)+f(b)+f(c)=0$ we obtain:$$a^3+b^3+c^3+q(a+b+c)-3r=0$$ whence $$a^3+b^3+c^3=3abc$$

Now let $a=\sqrt[3]{x+y}, b=\sqrt[3]{x-y}, c=-1$ to obtain:$$2x-1=-3\sqrt[3]{x^2-y^2}$$ You can then cube this and isolate $y$ to solve, which should give the answer you are looking for.

Note this method does not make it obvious that $x=-1$ is impossible.

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This is indeed a very elegant way of solving this problem, but (I think?) not a very intuitive approach. Thank you anyway! – flawr May 17 '14 at 21:57
@flawr I put it in because if you are encountering cube roots in this way (and this is related to the solution of cubic equations), then if $a+b+c=0$ you have the identity $a^3+b^3+c^3=3abc$. And this is occasionally a useful thing to know. – Mark Bennet May 18 '14 at 5:58
I was not aware of that, thank you for pointing this out! – flawr May 18 '14 at 18:57

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