Ta có: \(P=\frac{-\frac56ax^6y^3}{\left(2b^{2020}+8\right)^4}\)
=>\(P^2=\left(\frac{-\frac59a}{\left(2b^{2020}+8\right)^4}\right)^2\cdot\left(x^6y^3\right)^2=\frac{\frac{25}{9}a^2}{\left(2b^{2020}+8\right)^8}\cdot x^{12}y^6\)
=>Bậc của \(P^2\) là 12+6=18
