Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question

2 Answers 2

up vote 3 down vote accepted

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

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

$$2x+3(\sqrt[3]{x+y}\cdot\sqrt[3]{x-y})=1$$

$$3(\sqrt[3]{x+y}\cdot\sqrt[3]{x-y})=1-2x$$

$$3(\sqrt[3]{x^2-y^2})=1-2x$$

$$27(x^2-y^2)=(1-2x)^3$$

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

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

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

share|improve this answer
    
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 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.

share|improve this answer
    
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 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 at 5:58
    
I was not aware of that, thank you for pointing this out! –  flawr May 18 at 18:57

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.