\(a,=\left(3+x\right)\left(9-3x+x^2\right)\\ b,=\left(4x+0,1\right)\left(16x^2-0,4x+0,01\right)\\ c,=\left(2-3x\right)\left(4+6x+9x^2\right)\\ d,=\left(\dfrac{x}{5}-\dfrac{y}{3}\right)\left(\dfrac{x^2}{25}+\dfrac{xy}{15}+\dfrac{y^2}{9}\right)\)
a) \(27+x^3=3^3+x^3=\left(3+x\right)\left(9-3x+x^2\right)\)
b) \(64x^3+0,001=\left(4x\right)^3+\left(\dfrac{1}{10}\right)^3=\left(4x+\dfrac{1}{10}\right)\left(16x^2-\dfrac{4x}{10}+\dfrac{1}{100}\right)\)
a/\(27+x^3=\left(3+x\right)\left(9-3x+x^2\right)\)
b/ \(64x^3+0,001=\left(4x+0,1\right)\left(16x^2-0,4x+0,01\right)\)
c/ \(8-27x^3=\left(2-3x\right)\left(4+6x+9x^2\right)\)
d/ \(\dfrac{x^3}{125}-\dfrac{y^3}{27}=\left(\dfrac{x}{5}-\dfrac{y}{3}\right)\left(\dfrac{x^2}{25}+\dfrac{xy}{15}+\dfrac{y^2}{9}\right)\)