a) \(3\left(x+4\right)-x^2-4x\)

\(=3\left(x+4\right)-x\left(x+4\right)\)

\(=\left(3-x\right)\left(x+4\right)\)

b) \(x^2-2xy+y^2-z^2\)

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

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

ta có:

\(\left(2x-3\right)\left(x-2\right)\)

\(=2x^2-4x-3x-6\)

\(=2x^2-7x-6\)

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

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

\(=x^2+6x+9\)

\(\left(2x-3y\right)^3\)

\(=\left(2x\right)^3-3.\left(2x\right)^2.3y+3.2x.\left(3y\right)^2-\left(3y\right)^3\)

\(=8x^3-36x^2y+54xy^2-27y^3\)

\(2x\left(x^3-8\right)\)

\(=2x^4-16x\)

TL

(y-x) (z-x) (z-y)

HT

??????????????????????

Ta có x³ + y³ + z³ = 3xyz 

<=> x³ + y³ + z³ - 3xyz = 0

<=> (x + y)³ - 3xy(x + y) + z³ - 3xyz = 0

<=> [(x + y)³ + z³] - 3xy(x + y + z) = 0

<=> (x + y + z)[(x + y)² - (x + y)z + z²] - 3xy(x + y + z)  = 0

<=> (x + y + z)(x² + y² + z² + 2xy - xz - yz) - 3xy(x + y + z) = 0

<=> (x + y + z)(x² + y² + z² - xy - xz - yz) = 0

<=> \(\orbr{\begin{cases}x+y+z=0\\x^2+y^2+z^2-xy-yz-zx=0\end{cases}}\)(1) 

Khi x² + y² + z² - xy - xz - yz = 0

<=> 2(x² + y² + z² - xy - xz - yz) = 0

<=> 2x² + 2y² + 2z² - 2xy - 2xz - 2yz = 0

<=> (x² - 2xy + y²) + (y² - 2yz + z²) + (x² - 2xz + z²) = 0

<=> (x - y)² + (y - z)² + (x - z)² = 0

<=> \(\hept{\begin{cases}x-y=0\\y-z=0\\z-x=0\end{cases}}\Leftrightarrow x=y=z\)(2) 

Từ (2) (1) => đpcm