I want to find all the rational solutions of $x^2+y^3=z^3$ I want to find all the rational  solutions of $x^2+y^3=z^3$, 
and any help would be appreciated.
My answer is as follows：
$$x^2=z^3-y^3=(z-y)(z^2+zy+y^2)$$
Let $x=ab,a,b\in\Bbb Q$
when
$$ z-y=a^2,z^2+zy+y^2=b^2$$
$$z^2+z(z-a^2)+(z-a^2)^2=b^2$$
Sort it out, and that's a Pell's equation:
$$\left(\frac{2b}{a^2}\right)^2-3\left(\frac{2z}{a^2}-1\right)^2=1$$
Let $s=\frac{2b}{a^2},t=\frac{2z}{a^2}-1$
We obtained
$$s^2-3t^2=1$$
The least $s_0=2,t_0=1$,other solutions can be obtained from:
$$s_{n+1}=2s_n+3t_n,t_{n+1}=s_n+2t_n(n=0,1,2,...)$$
Last we obtained:
$$x=ab=\frac{a^3}{2}s_n,z=\frac{a^2}{2}(t_n+1),y=\frac{a^2}{2}(t_n-1)$$
Example:Let $a=\frac{1}{3},s_1=7,t_1=4$,then $x=\frac{7}{54},y=\frac{1}{6},z=\frac{5}{18}$.
I don't know what the other solutions are,I'd be glad of some help with this,thank you for you help.

EDIT: Let's first consider the more symmetric equation 
$$x^2=y^3+z^3\tag1.$$If $y=0$, we see that gives the solutions $$(x,y,z)=(t^3,0,t^2)$$ for rational $t$. If $y\neq0$, we can set $x=ry$ and $z=ty$ with rational $r,t$. That gives $r^2=y(1+t^3)$, i.e. $r=0$, giving $t=-1$ and thus the solutions $$(x,y,z)=(0,y,-y),$$ or $y=r^2/(1+t^3)$, giving the solutions $$(x,y,z)=\left(\frac{r^3}{1+t^3},\frac{r^2}{1+t^3},\frac{r^2t}{1+t^3}\right).$$
That should take care of all possible cases.
To return from (1) to the original equation, we just have to change the sign of $y$, so the original equation has the solutions
$$(x,y,z)=(t^3,0,t^2),$$
$$(x,y,z)=(0,y,y),$$
$$(x,y,z)=\left(\frac{r^3}{1+t^3},-\frac{r^2}{1+t^3},\frac{r^2t}{1+t^3}\right).$$
