a) (x-3).((3+x)2-(x2+3x+9))
<=> (x-3).(3x+9)
<=> 3.(x-3)2
a) đa thức
\(=\left(x-3\right)\left(x+3\right)^2-\left(x-3\right)\left(x^2+3x+9\right)=\left(x-3\right)\left(x^2+6x+9-x^2-3x-9\right)=3x\left(x-3\right)=3x^2-9x\)
b) đa thức \(=\left(x+6\right)^2-2x\left(x+6\right)+x^2-36=\left(x+6-x\right)^2-36=0\)