\(A=27x^3-27x^2+18x-6=3\left(9x^3-9x^2+6x-2\right)\)

\(B=2x^3-x^2+5x+6=2x^3-2x^2+x^2-x+6x+6==\left(2x^2+x+6\right)\left(x-1\right)\)

\(2x^2+2y^2+z^2+25-6y-2xy-8x+2z\left(y-x\right)=0\)

\(\Leftrightarrow2x^2+2y^2+z^2+25-6y-2xy-8x-2z\left(x-y\right)=0\)

\(\Leftrightarrow\left[x^2+y^2+z^2-2xy-2z\left(x-y\right)\right]+\left(x^2-8x+16\right)+\left(y^2-6y+9\right)=0\)

\(\Leftrightarrow\left[\left(x^2-2xy+y^2\right)-2z\left(x-y\right)+z^2\right]+\left(x-4\right)^2+\left(y-3\right)^2=0\)

\(\Leftrightarrow\left[\left(x-y\right)^2-2z\left(x-y\right)+z^2\right]+\left(x-4\right)^2+\left(y-3\right)^2=0\)

\(\Leftrightarrow\left(x-y-z\right)^2+\left(x-4\right)^2+\left(y-3\right)^2=0\)

Vì \(\left(x-y-z\right)^2\ge0\), \(\left(x-4\right)^2\ge0\), \(\left(y-3\right)^2\ge0\)\(\forall x,y,z\)

\(\Rightarrow\left(x-y-z\right)^2+\left(x-4\right)^2+\left(y-3\right)^2\ge0\)\(\forall x,y,z\)

Dấu " = " xảy ra \(\Leftrightarrow\hept{\begin{cases}x-y-z=0\\x-4=0\\y-3=0\end{cases}}\Leftrightarrow\hept{\begin{cases}z=x-y\\x=4\\y=3\end{cases}}\Leftrightarrow\hept{\begin{cases}z=4-3=1\\x=4\\y=3\end{cases}}\)

Vậy \(x=4\), \(y=3\), \(z=1\)

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

\(=\left(x+2-y\right)\left(x+2+y\right)\)

\(\frac{y^4-2y^2-8}{y-2}=\frac{\left(y^2-4\right)\left(y^2+2\right)}{y-2}=\frac{\left(y-2\right)\left(y+2\right)\left(y^2+2\right)}{y-2}\)

\(=\left(y+2\right)\left(y^2+2\right)=0\)

TH1 : \(y=-2\)

TH2 : vô nghiệm 

Vậy y = -2 

( y⁴ - 2y² - 8 ) : ( y - 2 ) = 0

<=> [ ( y⁴ - 2y² + 1 ) - 9 ] : ( y - 2 ) = 0

<=> [ ( y² - 1 )² - 3² ] : ( y - 2 ) = 0

<=> ( y² - 1 - 3 )( y² - 1 + 3 ) : ( y - 2 ) = 0

<=> ( y² - 4 )( y² + 2 ) : ( y - 2 ) = 0

<=> ( y - 2 )( y + 2 )( y² + 2 ) : ( y - 2 ) = 0

<=> ( y + 2 )( y² + 2 ) = 0

<=> y + 2 = 0 hoặc y² + 2 = 0

<=> y = -2 ( y² + 2 ≥ 2 > 0 ∀ y )

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

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

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

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

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

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