How Isolate y from $a=\sqrt{(y^2+(a+x)^2)^3}$

Question

Just that! I'm asking for a method to isolate $y$ from such expressions:

$$a=\sqrt{(y^2+(a+x)^2)^3}$$

or even easier:

$$a=\sqrt{(y^2+x^2)^3}.$$

EDIT:

I'm just trying to revisit some of the basics in algebraic equation manipulation. Even it's sooo easy to find the school examples with squares or cubes, dealing with radicals can twist into strange paths. The above expression ($a=\sqrt{(y^2+x^2)^3}$) can turn into a monster just as you add another simple term:

$$a=\sqrt{(y^2+x^2)^3} + \sqrt{(y^2-x^2)^3}$$

I can not see now how squaring the two sides could help in y isolation...

Answer 1

It might help to see how $\sqrt{(y^2+(a+x)^2)^3}$ is built starting from y

$\begin{align} & y \\ & {{()}^{2}}\to {{y}^{2}} \\ & +{{(a+x)}^{2}}\to {{y}^{2}}+{{(a+x)}^{2}} \\ & {{()}^{3}}\to {{({{y}^{2}}+{{(a+x)}^{2}})}^{3}} \\ & \sqrt{{}}\to \sqrt{{{({{y}^{2}}+{{(a+x)}^{2}})}^{3}}} \\ & =a \\ \end{align}$

So starting from y you do the sequence of operations $$()^2 \to +{{(a+x)}^{2}}\to ()^3 \to \sqrt{{}}$$ to produce a. You can recover y by starting from a and undoing those operations in reverse order $$()^2 \to ()^{\frac{1}{3}} \to -{{(a+x)}^{2}} \to \sqrt{{}}$$ to give

$\begin{align} & a \\ & ()^2 \to a^{2} \\ & ()^{\frac{1}{3}}\to a^{\frac{2}{3}} \\ & -(a+x)^2\to a^{\frac{2}{3}}-{(a+x)}^{2} \\ & \sqrt{}\to \pm \sqrt{a^{\frac{2}{3}}-(a+x)^{2}} \\ & = y \\ \end{align}$

Parsing the original expression this way can help you to see how to unravel it.

Answer 2

Taking the square of both sides of $a=\sqrt{(y^2+(a+x)^2)^3}$, one may get $$ a^2=(y^2+(a+x)^2)^3$$ Now we apply the cube root, $$ a^\frac{2}{3}=y^2+(a+x)^2$$ Then, we have $$ y^2=a^\frac{2}{3}-(a+x)^2$$ Finally, we take the square root of both sides, $$y=\pm \sqrt{a^\frac{2}{3}-(a+x)^2} $$