\(\left(x^2-1\right)^3-\left(x^4+x^2+1\right)\left(x^2-1\right)=0\)

\(\Leftrightarrow x^5-3x^4+3x^2-1-\left(x^6-x^4+x^4-x^2+x^2-1\right)=0\)

\(\Leftrightarrow x^5-3x^4+3x^2-x^6=0\)

\(\Leftrightarrow x^5\left(1-x\right)-3x^2\left(x^2-1\right)=0\Leftrightarrow x^5\left(1-x\right)-3x^2\left(x-1\right)\left(x+1\right)=0\)

\(\Leftrightarrow-x^5\left(x-1\right)-3x^2\left(x+1\right)\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left[-x^5-3x^3-3x^2\ne0\right]=0\Leftrightarrow x=1\)

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

\(\Leftrightarrow\left(x-3\right)\left(x^2+1\ne0\right)=0\Leftrightarrow x=3\)

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

\(\Rightarrow\)\(x^2.\left(x-3\right)+\left(x-3\right)=0\)

                  \(\left(x^2+1\right).\left(x-3\right)=0\)\(\Rightarrow\orbr{\begin{cases}x^2+1\\x-3\end{cases}}=0\)

Với :  \(x^2+1=0\Rightarrow x=\varnothing\)nhưng giá trị này làm cho biểu thức không có nghĩa, loại

         \(x-3=0\Rightarrow x=3\)

Vậy  \(x=3\)

a, Ta có : \(M=x^2-x=0\Leftrightarrow x\left(x-1\right)=0\Leftrightarrow x=0;1\)

b, Ta có : \(M=N\) hay \(x^2-x=\left(x-1\right)^3-x^2\left(x-3\right)-2\)

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

\(x^2-x=3x-3\Leftrightarrow x^2-4x+3=0\)

\(\Leftrightarrow\left(x-3\right)\left(x-1\right)=0\Leftrightarrow x=3;1\)

\(a^3-8a^2+16a=a\left(a^2-2.4a+4^2\right)=a\left(a-4\right)^2\)

\(x^3+2x^2-4x-8=x^2\left(x+2\right)-4\left(x+2\right)=\left(x^2-4\right)\left(x+2\right)=\left(x-2\right)\left(x+2\right)^2\)

\(A=\left(x-2\right)^3-x\left(x-1\right)\left(x-3\right)+3x^2-9x+8\) 

\(A=x^3-6x^2+12x-8-x^2+x^2-3x+3x^2-9x+8\)

\(A=x^3-6x^2+12x-3x+3x^2-9x\)

\(A=\left(-6x^2+3x^2\right)+\left(12x-3x-9x\right)+x^3\)

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

a) \(x^3+2x^2y+xy^2-4x=x\left(x^2+2xy+y^2-4\right)=x\left[\left(x+y\right)^2-2^2\right]=x\left(x+y+2\right)\left(x+y-2\right)\)

b) \(x^2-5x-y^2-5y=\left(x^2-y^2\right)-5\left(x+y\right)=\left(x-y\right)\left(x+y\right)-5\left(x+y\right)=\left(x-y-5\right)\left(x+y\right)\)