Question

[Băng Hắc Hường🖤💗]

 x^3 + 1=

8 + x^3=

27x^3 - 64y^3=

x^3 phần 64 - 1 phần 25

Answer 1

[HT.Phong (9A5)]

\(x^3+1\)

\(=x^3+1^3\)

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

______

\(8+x^3\)

\(=2^3+x^3\)

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

______

\(27x^3-64y^3\)

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

\(=\left(3x-4y\right)\left(9x^2+12xy+16y^2\right)\)

______

\(\dfrac{x^3}{64}-\dfrac{1}{125}\)

\(=\left(\dfrac{x}{4}\right)^3-\left(\dfrac{1}{5}\right)^3\)

\(=\left(\dfrac{x}{4}-\dfrac{1}{5}\right)\left(\dfrac{x^2}{16}+\dfrac{x}{20}+\dfrac{1}{25}\right)\)

Answer 2

[Nguyễn Lê Phước Thịnh]

x^3+1=(x+1)(x^2-x+1)

x^3+8=(x+2)(x^2-2x+4)

27x^3-64y^3=(3x-4y)(9x^2+12xy+16y^2)

\(\dfrac{x^3}{64}-\dfrac{1}{25}=\left(\dfrac{1}{4}x-\sqrt[3]{\dfrac{1}{5}}\right)\left(\dfrac{1}{16}x^2+\dfrac{1}{4\sqrt[3]{5}}\cdot x+\dfrac{1}{\sqrt[3]{25}}\right)\)