How can I isolate $y$ in $125=\left({\sqrt{y^2+\left(\frac{x+5}{2}\right)^2}}\right)^3+\left({\sqrt{y^2+\left (\frac{x-5}{2}\right)^2}}\right)^3$? I'm trying to find a method to isolate $y$ in this formula that I could use for any other numerical magnitudes or even positions when radicals are involved in this way
$$125=\left({\sqrt{y^2+\left(\frac{x+5}{2}\right)^2}}\right)^3+\left({\sqrt{y^2+\left (\frac{x-5}{2}\right)^2}}\right)^3$$
 A: I should not try to isolate $y$ since this would give a monster.
The first thing I should do is to replace $y^2=Y$ and consider that I look for the zero of equation $$f(Y)=\left({\sqrt{Y+\left(\frac{x+5}{2}\right)^2}}\right)^3+\left({\sqrt{Y+\left (\frac{x-5}{2}\right)^2}}\right)^3-125$$
Thsi equation does not show real root except for the range $-5\leq x \leq 5$ and $Y(-x)=Y(x)$. Replacing $x$ by a number leads to $Y$ and $Y(x)$ is a nice smooth curve. So, my idea would be


*

*Generate a table of values for $Y(x)$,as for example
$$\left(
\begin{array}{cc}
 x & Y \\
 0.0 & 9.49901 \\
 0.5 & 9.41167 \\
 1.0 & 9.14917 \\
1.5 & 8.71001 \\
 2.0 & 8.09152 \\
 2.5 & 7.28945 \\
 3.0 & 6.29726 \\
3.5 & 5.10467 \\
 4.0 & 3.69431 \\
4.5 & 2.03175 \\
 5.0 & 0.
\end{array}
\right)$$

*Build a cubic spline which, for a given $x$, will give you a good estimate $Y_0$ of the solution. Otherwise, use as an approximation $$ Y_0=\frac{2 \sqrt[3]{2}-1}{4} (5-x)(5+x)$$ In the worst case, a table lookup and linear interpolation would provide areasonable estimate of $Y$.

*Polish the solution using Newton method starting with $Y_0$ (notice that $f'(Y)$ is quite simple). The method would converge in very very few iterations.


For illustration purposes, let us consider $x=1.2345$; the approximate parabola gives an estimate $Y_0=9.1034$. Newton method generates the following iterates $$Y_1=8.96564$$ $$Y_2=8.96533$$ which is the solution for six significant figures.
A: Let $x = 10u$, $y = 5v$ and $w = u^2+v^2+\frac14$. The equation at hand can be rewritten as
$$\sqrt{v^2+\left(u+\frac12\right)^2}^3 + \sqrt{v^2+\left(u-\frac12\right)^2}^3 = 1 \iff \sqrt{w+u}^3 + \sqrt{w-u}^3 = 1$$
Let $f_{ab} = a\sqrt{w+u}^3 + b\sqrt{w-u}^3 - 1$ where $a, b \in \{ +, - \}$. 
Above equations becomes
$$f_{++} = 0 \quad\implies\quad f_{++}f_{+-}f_{-+}f_{--} = 0$$
When one expand LHS of last expression, all square roots cancel out. One get
$$36u^2w^4 - 4w^3 + 24u^4w^2 - 12u^2w + 4u^6 + 1 = 0$$
For fixed $x$, $u$ will be constant. Above equation becomes a quartic equation in $w$.
One can use formula discussed here to express $w$ in radicals.
Since $y = \pm 5\sqrt{ w - u^2 - \frac14}$,
one can express $y$ in radicals too. 
The problem is the formula for $y$ is a real mess (too horrible to reproduce here).
I doubt it has any practical use.
