a) \(-y^2+\dfrac{1}{9}\)

\(=-\left(y^2-\left(\dfrac{1}{3}\right)^2\right)\)

\(=-\left(y+\dfrac{1}{3}\right)\left(y-\dfrac{1}{3}\right)\)

b) \(4^4-256\)

\(=4^4-4^4\)

\(=0\)

c) \(9\left(x-3\right)^2-4\left(x+1\right)^2\)

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

\(=\left(3x-9+2x+2\right)\left(3x-9-2x-2\right)\)

\(=\left(5x-7\right)\left(x-11\right)\)

\(a,=\left(\dfrac{1}{3}-y\right)\left(\dfrac{1}{3}+y\right)\\ b,=\left(x^2-16\right)\left(x^2+16\right)\\ =\left(x-4\right)\left(x+4\right)\left(x^2+16\right)\\ c,=\left[3\left(x-3\right)-2\left(x+1\right)\right]\left[3\left(x-3\right)+2\left(x+1\right)\right]\\ =\left(3x-9-2x-2\right)\left(3x-9+2x+2\right)\\ =\left(x-11\right)\left(5x-7\right)\\ d,=\left(5x-\dfrac{1}{9}xy\right)\left(5x+\dfrac{1}{9}xy\right)=x^2\left(5-\dfrac{1}{9}y\right)\left(5+\dfrac{1}{9}y\right)\)

p =x(x³ - 64) = x(x-4)( x² +4x +16)

x^2-4+4xy-8y=x^2+4xy+4y^2-4y^2-8y-4=(x+2y)^2-(2y+2)^2=(x+2y-2y+2)(x+2y+2y-2)=(x+2)(x+4y-2)

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

\(9-x^2+2xy-y^2\)

\(=9-\left(x^2-2xy+y^2\right)\)

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

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

ta có : x^8 +x^4 +1= (x^8 -x^5) +(x^5-x^2) +(x^4 -x) +(x^2 +x 1)=x^5.(x^3 -1) +x^2(x^3-1) +x(x^3-1) +(x^2 +x+1)=x^5.(x-1)(x^2 +x+1) +x^2(x-1)(x^2 +x+1) +x(x-1)(x^2 +x+1) +(x^2 +x+10=(x^2 +x+1)(x^6- x^5 +x^3 -x +1)

Thay `x = 2` ta được :

`x^4+x^3-9x^2+10x-8`

`= 2^4 + 2^3 - 9*2^2 + 10*2 - 8`

`= 16 + 8 - 36 + 20 - 8`

`= 0`

Vậy `x = 2` là nghiệm của phương trình trên 

Do đó ta thực hiện phép chia :

\(\left(x^4+x^3-9x^2+10x-8\right):\left(x-2\right)\)

Vậy \(x^4+x^3-9x^2+10x-8=\left(x-2\right)\left(x^3+3x^2-3x+4\right)\).