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
 A: 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: $y^2 - x^3 < 0$ or $x^2 + y^3 < 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.

A: Rather than use $x$ directly, we will use $z=-x$. 
Suppose that $y>0$. Let 
$$
f_1(z)=z+\sqrt{y^2+z^3} \ (\text{for} \ z\geq{-y^{\frac{2}{3}}}), \ \
f_2(z)=\frac{y^3}{-y+\sqrt{y^3+z^2}} 
\ (\text{for} \ z\in{\mathbb R}), \
$$
We must then show that $f_1(z) \neq f_2(z)$ when $z\neq y$.
It is easy to see that $f_1$ is increasing on its domain and
that $f_2$ is increasing on $(-\infty,0)$ and decreasing on
$(0,\infty)$. Let $c=f_1(y)=f_2(y)$. 
When $z > y$, we have $f_2(z)<c<f_1(z)$ by the variations
of $f_1$ and $f_2$ so $f_1(z) \neq f_2(z)$. 
When $0 < z < y$, we have 
$f_1(z)<c<f_2(z)$ by the variations of $f_1$ and $f_2$ so $f_1(z) \neq f_2(z)$. 
When $z <0$, for $f_1(z)$ to be defined we need $z\geq \mu$ where
$\mu=-y^{\frac{2}{3}}$. Now $\mu^2=y^{\frac{4}{3}} \leq y^2$, so
$f_2(\mu) \geq f_2(y)=c$. As $f_1(z) <  c$ by the variations of $f_1$, we still
have $f_1(z)<c<f_2(z)$ so $f_1(z) \neq f_2(z)$.
The case $y < 0$ is similar.
