\(x^3+\frac{1}{27}\)
\(=x^3+\left(\frac{1}{3}\right)^3\)
\(=\left(x+\frac{1}{3}\right)\left(x^2-\frac{1}{3}x+\left(\frac{1}{3}\right)^2\right)\)
\(=\left(x+\frac{1}{3}\right)\left(x^2-\frac{1}{3}x+\frac{1}{9}\right)\)
Trả lời hoài mà chẳng ai T bất công thiệt
\(x^3+\frac{1}{27}=\left(x+\frac{1}{3}\right)\left(x^2-\frac{1}{3}x+\frac{1}{9}\right)\)
(Nếu đúng thì clik cho mình với nhoa!)
\(x^3+\frac{1}{27}=x^3+\left(\frac{1}{3}\right)^3=\left(x+\frac{1}{3}\right)\left[x^2-x.\frac{1}{3}+\left(\frac{1}{3}\right)^2\right]=\left(x+\frac{1}{3}\right)\left(x^2+\frac{1}{3}x+\frac{1}{9}\right)\)
\(x^3+\frac{1}{27}\)
\(=x^3+\left(\frac{1}{3}\right)^3\)
\(=\left(x+\frac{1}{3}\right).\left(x^2-\frac{1}{3}.x+\frac{1}{9}\right)\)