\(\left(x+\frac{-2}{9}\right)^3=\left(\frac{8}{27}\right)^2\\ \left(x+\frac{-2}{9}\right)^3=\left[\left(\frac{2}{3}\right)^3\right]^2\\ \left(x+\frac{-2}{9}\right)^3=\left(\frac{2}{3}\right)^{3\cdot2}\\ \left(x+\frac{-2}{9}\right)^3=\left[\left(\frac{2}{3}\right)^2\right]^3\\ \left(x+\frac{-2}{9}\right)^3=\left(\frac{4}{9}\right)^3\\ \Rightarrow x+\frac{-2}{9}=\frac{4}{9}\\ x=\frac{4}{9}+\frac{2}{9}\\ x=\frac{6}{9}=\frac{2}{3}\)
Vậy \(x=\frac{2}{3}\)