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I am working on a physics task, and reduced it to the following equation for $y$: $$\frac{1}{4y^3}-\frac{2}{(y^2+b^2)^{\frac{3}{2}}}=0$$ I handed it to Mathematica, and it gave me two real solutions, $$y_{1,2} = \pm\frac{b}{\sqrt{3}},$$ along with some complex ones. My question is, how can I see this? I mean, how can I solve such an equation by hand? I tried moving stuff around a bit, but all I found was a sextic equation, which I really didn't feel like approaching. Is there any easier way? Does it lower the hassle, if only real solutions are to be found?

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  • $\begingroup$ depending on the context there are 6 solutions. (the 3 complex roots of $y^{3}$ for each of the solutions you gave) $\endgroup$
    – Max
    Feb 28, 2014 at 19:27
  • $\begingroup$ $y$ is to be interpreted as a radius, so actually the positive solution suffices. $\endgroup$ Feb 28, 2014 at 19:38
  • $\begingroup$ Any equation that can be solved by any method can be solved by hand. It just might take a long time. $\endgroup$ Mar 1, 2014 at 17:05

3 Answers 3

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First, get rid of the denominators,

$$\begin{align} \frac{1}{4y^3} - \frac{2}{(y^2+b^2)^{3/2}} &= 0\\ \iff \frac{1}{4y^3} &= \frac{2}{(y^2+b^2)^{3/2}}\\ \iff (y^2+b^2)^{3/2} &= 8y^3. \end{align}$$

Then, raise it to the $2/3$-th power to simplify it, introducing a third root of unity,

$$\begin{align} y^2 + b^2 &= 4\rho y^2\\ \iff b^2 &= (4\rho-1)y^2\\ \iff y &= \pm \frac{b}{\sqrt{4\rho-1}}. \end{align}$$

Choosing $\rho = 1$ as the third root of unity yields the two real solutions. $\rho = e^{\pm 2\pi i/3}$ yields non-real solutions.

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Let $x^2=y^2+b^2$. Then $x=\pm2y$ and you find the result

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    $\begingroup$ This answer gives a "magic" solution, rather than explaining how, which is what the question asked. $\endgroup$
    – Brian S
    Feb 28, 2014 at 20:24
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    $\begingroup$ I didn't think it was magic, as it a way to get rid of the cubes. It might be easier to find by clearing fractions and taking the cube root of both sides. $\endgroup$ Feb 28, 2014 at 21:04
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Hint

Rearrange terms, so you can get $$ 8y^3 - \left( \sqrt{y^2+b^2} \right )^3 = 0 $$ Now, use the fact that $a^3 - b^3 = (a-b)(a^2+ab+b^2)$ $$ \left( 2y - \sqrt{y^2+b^2} \right )\left( 4y^2 + 2y \sqrt{y^2+b^2} + y^2 + b^2\right ) = 0 $$ Can you take it from here?

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    $\begingroup$ $a^3-b^3=(a-b)(a^2+ab+b^2)$ should be more popular indeed! $\endgroup$ Mar 1, 2014 at 9:56

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