\(\left(\frac{1-\sqrt{a}}{1-a}\right)^2\)

=\(\left[\frac{1-\sqrt{a}}{\left(1-\sqrt{a}\right)\left(1+\sqrt{a}\right)}\right]^2\)

=\(\left(\frac{1}{1+\sqrt{a}}\right)^2\)

=\(\frac{1}{1+2\sqrt{a}+a}\)

\(\left(\frac{1-\sqrt{a}}{1-a}\right)^2\)

\(=\left[\frac{1-\sqrt{a}}{\left(1-\sqrt{a}\right)\left(1+\sqrt{a}\right)}\right]^2\)

\(=\left(\frac{1}{1+\sqrt{a}}\right)^2\)

\(=\frac{1}{1+2\sqrt{a}+a}\)

\(x^3+3x^2+3x+2\)

\(=\left(x+2\right)\left(x^2+x+1\right)\)

\(x^3+3x^2+3x+2\)

\(=x^3+2x^2+x^2+2x+x+2\)

\(=x^2\left(x+2\right)+x\left(x+2\right)+\left(x+2\right)\)

\(=\left(x+2\right)\left(x^2+x+1\right)\)

\(x^3+3x^2 +3x+2\)

\(=x^3+x^2+x+2x^2+2x+2\)

\(=x\left(x^2+x+1\right)+2\left(x^2+x+1\right)\)

\(=\left(x+2\right)\left(x^2+x+1\right)\)

Gọi thương khi chia R(x) cho \(x^2+x-2\) ,ta có: 

\(R\left(x\right)=x^3+ax+b=\left(x^2+x-2\right)Q\left(x\right)=\left(x-1\right)\left(x+2\right)Q\left(x\right)\)

Cho lần lượt \(x=1,x=-2\) , ta được: 

            \(\hept{\begin{cases}1^3+a.1+b=0\\\left(-2\right)^3-2a+b=0\end{cases}\Rightarrow\hept{\begin{cases}1+a+b=0\\-8-2a+b=0\end{cases}\Rightarrow}\hept{\begin{cases}a+b=-1\\-2a+b=8\end{cases}}}\)

          \(\Rightarrow\hept{\begin{cases}a+b=-1\\-2a+b-\left(a+b\right)=8-\left(-1\right)\end{cases}}\Rightarrow\hept{\begin{cases}a+b=-1\\-3a=9\end{cases}\Rightarrow\hept{\begin{cases}-3+b=-1\\a=-3\end{cases}}}\)

Vậy \(a=-3,b=2\)

\(x^3+3x^2+3x+2\)

\(=x^3+2x^2+x^2+2x+x+2\)

\(=x^2\left(x+2\right)+x\left(x+2\right)+\left(x+2\right)\)

\(=\left(x+2\right)\left(x^2+x+1\right)\)