\(a,=\left(x-2\right)\left(x^2+2x+4\right)\\ b,=10x\left(x-y\right)-5\left(x-y\right)=5\left(2x-1\right)\left(x-y\right)\\ c,=\left(x^2-4xy+4y^2\right)-9\\ =\left(x-2y\right)^2-3^2=\left(x-2y-3\right)\left(x-2y+3\right)\\ d,=x^2+3x-4x-12=x\left(x+3\right)-4\left(x+3\right)=\left(x-4\right)\left(x+3\right)\)
a) \(x^3-8=\left(x-2\right)\left(x^2+2x+4\right)\)
b) \(10x^2-10xy-5x+5y=10x\left(x-y\right)-5\left(x-y\right)=5\left(x-y\right)\left(2x-1\right)\)
c) \(x^2-9-4xy+4y^2=\left(x^2-4xy+4y^2\right)-9=\left(x-2y\right)^2-9=\left(x-2y-3\right)\left(x-2y+3\right)\)
d) \(x^2-x-12=\left(x^2-x+\dfrac{1}{4}\right)-\dfrac{49}{4}=\left(x-\dfrac{1}{2}\right)^2-\dfrac{49}{4}=\left(x-\dfrac{1}{2}-\dfrac{7}{2}\right)\left(x-\dfrac{1}{2}+\dfrac{7}{2}\right)=\left(x-4\right)\left(x+3\right)\)
