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

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

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

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

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

\(x^2+4y^2+3x-6y=\left(x^2+3x\right)-\left(4y^2+6y\right)=x\left(x+3\right)-2y\left(2y+3\right)\)

\(a,\left(3x+1\right)^2-\left(x+1\right)^2\)

\(=\left(3x+1-x-1\right)\left(3x+1+x+1\right)\)

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

\(=4x\left(2x+1\right)\)

\(b,6x-6y-x^2+xy\)

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

\(=6\left(x-y\right)-x\left(x-y\right)\)

\(=\left(x-y\right)\left(6-x\right)\)

a ) = 3 ( x^2 + 2xy + y^2 - z^2 )

     = 3  [ ( x  +  y)^2 - z^2] 

      = 3 ( x + y - z)( x  + y + z)

b) 4 y^ 2 - 5 y - 6 = 4y^2 - 8y + 3y - 6 = 4y ( y-  2 ) + 3 ( y-  2 ) = (  4y +3 )( y - 2 )

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

\(3x^4-48=3\left(x^4-16\right)=3\left[\left(x^2\right)^2-4^2\right]\\ =3\left(x^2-4\right)\left(x^2+4\right)\\ =3\left(x-2\right)\left(x+2\right)\left(x^2+4\right)\)

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

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

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