ta có 

\(x\left(x+4\right)\left(x+6\right)\left(x+10\right)+128=\left(x^2+10x\right)\left(x^2+10x+24\right)+128\)

\(=\left[\left(x^2+10x+12\right)-12\right]\left[\left(x^2+10x+12\right)+12\right]+128\)

\(=\left(x^2+10x+12\right)^2-12^2+128=\left(x^2+10x+12\right)^2-16\)

\(=\left(x^2+10x+12-4\right)\left(x^2+10x+12+4\right)=\left(x^2+10x+8\right)\left(x^2+10x+16\right)\)

\(=\left(x+2\right)\left(x+8\right)\left(x^2+10x+8\right)\)

Ta có: \(6x^4-13x^2+6\)

\(=6x^4-9x^2-4x^2+6\)

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

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

\(6x^4-13x^2+6=6x^4-4x^2-9x^2+6\)

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

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

Ta có:\(6x^4-13x^2+6\)

\(=6x^4-9x^2-4x^2+6\)

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

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

\(x^6-y^6\)

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

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

x^{6 -} y⁶

= ( x³ )² - ( y³ ) ²

⁼ ( x³ + y³) ( x³ - y³ )

= ( x - y ) ( x² + xy + y^{2 )}.( x + y ) ( x² - xy + y^{2 )}

am chiu cau nay kho qua em moi hoc lop 6

\(x^6-y^6=\left(x^3\right)^2-\left(y^3\right)^2=\left(x^3-y^3\right)\left(x^3+y^3\right)\)

\(x^6-y^6\)

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

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

\(x^6-y^6\\ =\left(x^3\right)^2-\left(y^3\right)^2\\ =\left(x^3-y^3\right)\left(x^3+y^3\right)\\ =\left(x-y\right)\left(x^2+xy+y^2\right)\left(x+y\right)\left(x^2-xy+y^2\right)\)

x⁶-y⁶ =(x-y)(x⁵+x⁴y+x³y²+x²y³+xy⁴+y⁵)

\(3x\cdot\left(x-y\right)^2-6\cdot\left(y-x\right)\)

\(=3x\left(x-y\right)^2+6\left(x-y\right)\)

\(=\left(x-y\right)\left[3x\left(x-y\right)+6\right]\)

\(=\left(x-y\right)\left(3x^2-3xy+6\right)\)

(x - y)(3x² - 3xy + 6)