\(1-3x-x^3+3x^2\)\(=\left(1-x^3\right)+\left(3x^2-3x\right)\)

\(=\left(1-x\right)\left(x^2+x+1\right)+3x\left(x-1\right)\)

\(=\left(x-1\right)\left(3x-x^2-x-1\right)=\left(x-1\right)\left(2x-x^2-1\right)\)

\(a,\left(3x+1\right)^2-\left(x+1\right)^2\)

\(=\left(3x+1-x-1\right)\left(3x+1+x+1\right)\)

\(=2x\left(4x+2\right)\)

\(=4x\left(2x+1\right)\)

\(b,6x-6y-x^2+xy\)

\(=\left(6x-6y\right)-\left(x^2-xy\right)\)

\(=6\left(x-y\right)-x\left(x-y\right)\)

\(=\left(x-y\right)\left(6-x\right)\)

x³-3x²-3x+1=x³+1-3x²-3x

                  =(x+1)(x²-x+1)-3x(x+1)

                  =(x+1)(x²-x+1-3x)

                  =(x+1)(x²-4x+1)

3x(x+1)²-5x²(x+1)+7(x+1)

=(x+1)(-5x²+3x+7)

x^5+2x^4+2x^3+2x^2+2x+1

=(x^5+x^4)+(x^4+x^3)+(x^3+x^2)+(x^2+x)+(x+1)

=x^4(x+1)+x^3(x+1)+x^2(x+1)+x(x+1)+(x+1)

=(x+1)(x^4+x^3+x^2+x+1)

Câu a :

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

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

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

\(=x\left(x-1\right)-2\left(x-1\right)\)

\(=\left(x-1\right)\left(x-2\right)\)

\(x^3-3x^2+3x-1\) =0

=>\(\left(x-1\right)^3\)=0

=>x-1=0

=>x=1

vậy x =1

\(x^3-3x^2+3x-1=0\)

\(\Leftrightarrow\left(x-1\right)^3\)

\(\Leftrightarrow x-1=0\)

\(\Leftrightarrow x=1\)

x²-4x+x+4

=x²+x-4x+4

=x(x+1)-4(x+1)

=(x+1)(x-4)

\(x^2-3x+4\)

\(=x^2+x-4x+4\)

\(=\left(x^2+x\right)-\left(4x+4\right)\)

\(=x\left(x+1\right)-4\left(x+1\right)\)

\(=\left(x-4\right)\left(x+1\right)\).