Đặt \(A=x\left(x-2\right)\left(x+a\right)\left(x+2a\right)\)
\(=x\left(x+a\right)\left(x-a\right)\left(x+2a\right)\)
\(=\left(x^2+ax\right)\left(x^2+ax-2a^2\right)\)
Đặt \(x^2+ax=t\)
\(\Rightarrow A=t\left(t-2a^2\right)\)
\(\Rightarrow\)\(x\left(x-2\right)\left(x+a\right)\left(x+2a\right)+a^4=t\left(t-2a^2\right)+a^4\)
\(=a^4-2a^2t+t^2=\left(a^2-t\right)^2=\left(a^2-x^2-ax\right)^2\)(là bình phương của 1 đa thức)