Đặt hay không đặt cũng như nhau bạn
\(\left(\frac{1-x^3}{1-x}+x\right)\left(\frac{1-x}{1-x^2}\right)^2=\left(\frac{\left(1-x\right)\left(1+x+x^2\right)}{1-x}+x\right)\left[\frac{1-x}{\left(1-x\right)\left(1+x\right)}\right]^2\)
\(=\left(1+2x+x^2\right)\left(\frac{1}{1+x}\right)^2=\left(1+x\right)^2\frac{1}{\left(1+x\right)^2}=1\)