Giải thích các bước giải:
Đặt $\dfrac{8x^3+2020}{2021}=t$
$\to\begin{cases}8x^3+2020=2021t\\ t^3=4042x-2020\end{cases}$
$\to\begin{cases}(2x)^3+2020=2021t\\ t^3+2020=2021.2x\end{cases}$
$\to (2x)^3+2020-(t^3+2020)=2021t-2020.2x$
$\to (2x)^3-t^3=2021(t-2x)$
$\to (2x-t)(4x^2+2xt+t^2)=-2021(2x-t)$
$\to (2x-t)(4x^2+2xt+t^2)+2021(2x-t))$
$\to (2x-t)(4x^2+2xt+t^2+2021)$
$\to 2x-t=0$ vì $4x^2+2xt+t^2+2021=(2x+\dfrac12t)^2+\dfrac34t^2+2021>0$
$\to 2x=t$
$\to 2x=\dfrac{8x^3+2020}{2021}$
$\to 8x^3-4042x+2020=0$
$\to 2\left(2x-1\right)\left(2x^2+x-1010\right)=0$
$\to x=\dfrac12$
