How prove this $x+y=0$ if $\left(\sqrt{y^2-x^3}-x\right)\left(\sqrt{x^2+y^3}-y\right)=y^3$ Question:

let $x,y$ are real numbers,and such


  $$\left(\sqrt{y^2-x^3}-x\right)\left(\sqrt{x^2+y^3}-y\right)=y^3$$
  show that
  $$x+y=0\tag{1}$$


before I have solve following problem:

if 


  $$(x+\sqrt{x^2+1})(y+\sqrt{y^2+1})=1,\Longrightarrow x+y=0\tag{2}$$
  solution:we have
  $$y+\sqrt{y^2+1}=\sqrt{x^2+1}-x\tag{3}$$
  $$x+\sqrt{x^2+1}=\sqrt{y^2+1}-y\tag{4}$$
  $(3)+(4)$
  then  $x+y=0$


But for $(1)$ we can't use this methods to solve  it.maybe can have other nice methods, Thank you
Below some characteristics of the function $f(x,y) = \left(\sqrt{y^2-x^3}-x\right)\left(\sqrt{x^2+y^3}-y\right)-y^3$ are shown. The area where $f(x,y)$ is undefined is in gray: $y^2 - x^3 &lt; 0$ or $x^2 + y^3 &lt; 0 $. Therefore the domain of the function is mainly restricted to $x\le 0$ and $y\ge 0$. The line $x+y=0$ is in red, some contour lines of $f(x,y)$ are in blue. The function is somewhat symmetrical around $x+y=0$, but not quite. Shouldn't the restriction $x\le 0$ and $y\ge 0$ be added to the problem in the first place?It's easy to prove that $x+y=0$ is a solution of $f(x,y)=0$, but proving the reverse seems to be tedious; at least I see no "nice" way to do it.
We'll replace $x$ on $-x$ and have the following problem.

Let $x$ and $y$ be real numbers such that $$\left(\sqrt{y^2+x^3}+x\right)\left(\sqrt{x^2+y^3}-y\right)=y^3.$$
Prove that
$$x=y.$$
Indeed, we need to prove that
$$\left(\sqrt{y^2+x^3}+x\right)\left(\sqrt{x^2+y^3}-y\right)=\left(\sqrt{x^2+y^3}+x\right)\left(\sqrt{x^2+y^3}-x\right)$$ or
$$\left(\sqrt{y^2+x^3}+x\right)\left(\sqrt{x^2+y^3}-y\right)-\left(\sqrt{x^2+y^3}+x\right)\left(\sqrt{x^2+y^3}-y\right)+$$
$$+\left(\sqrt{x^2+y^3}+x\right)\left(\sqrt{x^2+y^3}-y\right)-\left(\sqrt{x^2+y^3}+x\right)\left(\sqrt{x^2+y^3}-x\right)=0$$ or
$$\left(\sqrt{y^2+x^3}-\sqrt{x^2+y^3}\right)\left(\sqrt{x^2+y^3}-y\right)+(x-y)\left(\sqrt{x^2+y^3}+x\right)=0,$$ which gives $$x=y$$ or
$$\frac{(x^2+xy+y^2-x-y)\left(\sqrt{x^2+y^3}-y\right)}{\sqrt{y^2+x^3}+\sqrt{x^2+y^3}}+\sqrt{x^2+y^3}+x=0,$$ which is
$$\sqrt{(x^2+y^3)(y^2+x^3)}+x\sqrt{y^2+x^3}+(x^2+xy+y^2-y)\sqrt{x^2+y^3}+x^2+xy+y^2-x^2y-xy^2=0.$$
We'll prove that
$$\sqrt{(x^2+y^3)(y^2+x^3)}+x\sqrt{y^2+x^3}+(x^2+xy+y^2-y)\sqrt{x^2+y^3}+x^2+xy+y^2-x^2y-xy^2\geq0,$$ for which we consider four cases.


$x\geq0$ and $y\geq0.$


We see that $$\sqrt{(x^2+y^3)(y^2+x^3)}-y\sqrt{x^2+y^3}=\sqrt{x^2+y^3}\left(\sqrt{y^2+x^3}-y\right)\geq0$$ and
$$(x^2+xy+y^2)\sqrt{x^2+y^3}-x^2y-xy^2\geq x(x^2+xy+y^2)-x^2y-xy^2=x^3\geq0.$$
The equality occurs for $x=y=0$ only.


$x\geq0,$ $y\leq0$.


After replacing $y$ on $-y$ we need to prove that
$$\sqrt{(x^2-y^3)(y^2+x^3)}+x\sqrt{y^2+x^3}+(x^2-xy+y^2+y)\sqrt{x^2-y^3}+x^2-xy+y^2+x^2y-xy^2\geq0,$$ where $x$ and $y$ are non-negatives.

We see that $x^2\geq y^3$, which gives
$$x\sqrt{y^2+x^3}\geq x\sqrt{y^2+y^{4.5}}\geq xy^2.$$
The equality occurs only for $x=y=0$ again.


$x\leq0$ and $y\geq0.$


We'll replace $x$ on $-x$ and we need to prove that:
$$\sqrt{(x^2+y^3)(y^2-x^3)}-x\sqrt{y^2-x^3}+(x^2-xy+y^2-y)\sqrt{x^2+y^3}+x^2-xy+y^2-x^2y+xy^2\geq0,$$ 
where $x$ and $y$ are non-negatives now.

We see that
$$\sqrt{(x^2+y^3)(y^2-x^3)}-x\sqrt{y^2-x^3}=\sqrt{y^2-x^3}\left(\sqrt{x^2+y^3}-x\right)\geq0.$$
Thus, it's enough to prove that
$$(x^2-xy+y^2-y)\sqrt{x^2+y^3}+x^2-xy+y^2-x^2y+xy^2\geq0,$$ which is true, but my proof of this statement is still very ugly.


$x\le0$ and $y\leq0$.


After replacing $x$ on $-x$ and $y$ on $-y$ we need to prove that:

$$\sqrt{(x^2-y^3)(y^2-x^3)}-x\sqrt{y^2-x^3}+(x^2-xy+y^2+y)\sqrt{x^2-y^3}+x^2-xy+y^2+x^2y+xy^2\geq0,$$
where $x$ and $y$ are non-negative, for which it's enough to prove that
$$x^2-xy+y^2\geq x\sqrt{y^2-x^3}.$$
Now, by AM-GM $$x\sqrt{y^2-x^3}\leq\frac{1}{2}(x^2+y^2-x^3).$$
Id est, it's enough to prove that
$$x^2-xy+y^2\geq\frac{1}{2}(x^2+y^2-x^3)$$ or
$$(x-y)^2+x^3\geq0,$$ which is obvious.

The equality occurs for $x=y=0.$