\(2,\\ a,=\left(2x-1\right)^3\\ b,=x^3+3x^2y+3x^2y+9xy^2+3xy^2+9y^3\\ =x^2\left(x+3y\right)+3xy\left(x+3y\right)+3y^2\left(x+3y\right)\\ =\left(x^2+3xy+3y^2\right)\left(x+3y\right)\\ 3,\\ a,=x^2-2x+4x-8=\left(x-2\right)\left(x+4\right)\\ b,=x^2+2x+3x+6=\left(x+2\right)\left(x+3\right)\\ c,=4x^2-4x-8x+8=\left(x-1\right)\left(4x-8\right)=4\left(x-2\right)\left(x-1\right)\\ d,=x^2-xy+\dfrac{1}{4}y^2-y^2=\left(x-\dfrac{1}{2}y\right)^2-y^2\\ =\left(x-\dfrac{1}{2}y-y\right)\left(x-\dfrac{1}{2}y+y\right)=\left(x-\dfrac{3}{2}y\right)\left(x+\dfrac{1}{2}y\right)\)
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