\(3a^2c^2+bd+3abc+acd\)

\(=\left(3a^2c^2+3abc\right)+\left(acd+bd\right)\)

\(=3ac\left(ac+b\right)+d\left(ac+b\right)\)

\(=\left(ac+b\right)\left(3ac+d\right)\)

     \(-16x^2+8xy-y^2+49\)

\(=49-\left(16x^2-8xy+y^2\right)\)

\(=7^2-\left(4x-y\right)^2\)

\(=\left(7-4x+y\right)\left(7+4x-y\right)\)

      \(x^6-x^4+2x^3+2x^2\)

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

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

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

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

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

 \(\left(x+y+z\right)^2=x^2+y^2+z^2+2\left(xy+yz+xz\right)\)

Thay số vào tính được \(xy+yz+xz=12\)

Ta có: \(x^2+y^2+z^2=xy+yz+xz\left(=12\right)\)

\(\Rightarrow2x^2+2y^2+2z^2-2xy-2yz-2xz=0\) 

\(\Rightarrow\left(x^2-2xy+y^2\right)+\left(y^2-2yz+z^2\right)+\left(x^2-2xz+z^2\right)=0\)

\(\Rightarrow\left(x-y\right)^2+\left(y-z\right)^2+\left(x-z\right)^2=0\)

Từ đó được \(x=y=z\)

Mà \(x+y+z=6\Rightarrow x=y=z=2\)

Chúc bạn học tốt.

Ta có :

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

\(\Leftrightarrow ax^3+\left(b-a\right).x^2-\left(a+b\right).x-b\)

\(=ax^3+cx^2-1\)

\(\Leftrightarrow\hept{\begin{cases}b-a=c\\a+b=0\\b=1\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}a=-1\\b=1\\c=2\end{cases}}\)

Vậy ...

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

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

hay  \(A=x^2\left(x-2\right)+x\left(x-2\right)-6\left(x-2\right)\)

\(=\left(x-2\right)\left(x^2+x+6\right)\)

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

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

   \(=x^3-2x^2+x^2-2x-6x+12\)

   \(=x^2\left(x-2\right)+x\left(x-2\right)-6\left(x-2\right)\)

   \(=\left(x-2\right)\left(x^2+x-6\right)\)

   \(=\left(x-2\right)\left[x\left(x+3\right)-2\left(x+3\right)\right]\)

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

Chúc bạn học tốt.