Question

Giúp mình giải câu này với. Cảm ơn

Answer 1

\(a,=3^3+\left(2y\right)^3=\left(3+2y\right).\left(9-6y+4y^2\right)\\ b,=\left(\dfrac{1}{2}x\right)^3-1^3=\left(\dfrac{1}{2}x-1\right).\left(\dfrac{1}{4}x^2+\dfrac{1}{2}x+1\right)\\ c,=-\left(x^3+\dfrac{1}{27}\right)=-\left[x^3+\left(\dfrac{1}{3}\right)^3\right]=-\left(x+\dfrac{1}{3}\right).\left(x^2+\dfrac{1}{3}x+\dfrac{1}{9}\right)\)

Answer 2

\(a,\\ 27+8y^3\\ =\left(2y+3\right)\left(4y^2-6y+9\right)\\ b,\\ \dfrac{1}{8}x^3-1\\ =\dfrac{1}{8}\left(x-2\right)\left(x^2+2x+4\right)\\ c,\\ -x^3-\dfrac{1}{27}\\ =-\dfrac{1}{27}\left(27x^3+1\right)\\ =-\dfrac{1}{27}\left(3x+1\right)\left(9x^2-3x+1\right)\)

Answer 3

a,=33+(2y)3=(3+2y).(9−6y+4y2)b,=(\(\dfrac{1}{2}\)x)\(^3\)
−13=(\(\dfrac{1}{2}\)x−1).(\(\dfrac{1}{4}\)x2+\(\dfrac{1}{2}x\)+1)c,=−(x3+\(\dfrac{1}{27}\))=−[x3+(\(\dfrac{1}{3}\))\(^3\)
]=−(x+\(\dfrac{1}{3}\)).(x2+\(\dfrac{1}{3}\)x+\(\dfrac{1}{9}\))