Solve $\begin{cases} (x-2019)(2020+2021\sqrt[3]{y-2022})=1\\ \sqrt[3]{x-2019}(y-4042)=2021 \end{cases}$ Solve $$\begin{cases}
(x-2019)(2020+2021\sqrt[3]{y-2022})=1\\
\sqrt[3]{x-2019}(y-4042)=2021
\end{cases}$$

I tried to put $u=\sqrt[3]{x+2019}$ and $v=\sqrt[3]{y-2022}$ but not succed to find $u$ and $v$.
Any help is appreciated. Thank you.
$$y-4042=2021\sqrt[3]{2020+2021\sqrt[3]{y-2022}}$$ or
$$y-2022=2020+2021\sqrt[3]{2020+2021\sqrt[3]{y-2022}}.$$
Now, let $\sqrt[3]{y-2022}=t.$

Thus, $$\frac{t^3-2020}{2021}=\sqrt[3]{2020+2021t}$$ or
$$f(t)=g(t),$$ where $$f(t)=\frac{t^3-2020}{2021}$$ and $$g(t)=\sqrt[3]{2020+2021t}.$$
But easy to see that $f$ and $g$ are an inverse functions and $f$ increases.

Thus, $$f(t)=g(t)$$ is equivalent to $$f(t)=t$$ or
$$t^3-2021t-2020=0$$ or
$$t^3+t^2-t^2-t-2020t-2020=0$$ or
$$(t+1)(t^2-t-2020)=0.$$ 
Can you end it now?
With your substitution:

$$u = \sqrt[3]{x-2019},\ v=\sqrt[3]{y-2022}$$

we can substitute:

$$u = \frac{2021}{v^3-2020}$$

from the second equation into the first one:

$$\frac{2021^3(2020+2021v)}{(v^3-2020)^3}=1$$

or

$$2021^3[2021(v+1)-1]=(v^3-2020)^3$$

or

$$2021^4(v+1)=(v^3+1)[2021^2-2021(v^3-2020)+(v^3-2020)^2]$$

Clearly $v= -1$ is a solution and this corresponds to the solution $(x,y)=\color{green}{\boxed{(2018,2021)}}$. It remains to discuss over:

$$2021^4=(v^2-v+1)[2021^2-2021(v^3-2020)+(v^3-2020)^2]$$

By computer, this factors into:

$$(v^2-v-2020)(v^6+2021v^4-4040v^3+4084441v^2- 4082420 v+8258735661)=0$$

The first equation gives two other solutions: $v = \frac{1}{2}\pm \frac{\sqrt{8081}}{2}$ which correspond to:

$$(x,y)=\color{green}{\boxed{\left(2019+\frac{8}{(\sqrt{8081}+1)^3},\frac{2021}{2}(5+\sqrt{8081})\right)}}$$

$$(x,y)=\color{green}{\boxed{\left(2019-\frac{8}{(\sqrt{8081}-1)^3},\frac{2021}{2}(5-\sqrt{8081})\right)}}$$

For the second equation, we can prove by AM-GM that:

$$2020v^4+2020v^2 \geq 4040v^3$$

and for $$4082421v^2-4082420v+8258735661$$ we have $\Delta &lt; 0$, so the second factor is always positive.