Show that solution $(0,0)$ is unique Assume the system:
$$
\begin{align*}
& 2x-5y-x(x^2+2y^2)^2=0 \\
& 5x+2y-3y(2x^2+y^2)^2=0 
\end{align*}
$$
Obviously, $(0,0)$ is a solution. The thing I find hard doing is showing that it's the only solution.
Attempt:
Suppose that there exist another solution $(\bar{x},\bar{y}) \neq (0,0)$. Then, for $x \neq0$ and $y \neq 0$:
$$
2x-5y=x(x^2+2y^2)^2 \iff 2-5\frac{y}{x}=(x^2+2y^2)^2=c(x,y)\geq 0 \tag{1}
$$
$$
5x+2y=3y(2x^2+y^2)^2 \iff 5 \frac{x}{y}+2=3(2x^2+y^2)^2=k(x,y) \geq 0\tag{2}
$$
From equations $(1)$ and $(2)$, we yield:
$$
2-5\frac{5}{k-2}=c\iff (2-c)(k-2)=25
$$
But I can't seem to spot a contradiction. Is there  another (more straightforward) way to prove this?

If $y=0$ so $x=0$ and we get $(0,0)$.
Let $y\neq0$ and $x=ty$.
Thus, $$\frac{2t-5}{5t+2}=\frac{t(t^2+2)^2}{3(2t^2+1)^2}$$ or
$$5t^6-22t^5+80t^4-16t^3+80t^2+2t+15=0,$$ which is impossible for all real $t$.
Indeed,
$$5t^6-22t^5+80t^4-16t^3+80t^2+2t+15=$$
$$=(5t^6-22t^5+25t^4)+(55t^4-16t^3+40t^2)+(40t^2+2t+15)>0.$$
