\(3\left(3x-\dfrac{1}{2}\right)^3+\dfrac{1}{9}=0\)
=>\(3\left(3x-\dfrac{1}{2}\right)^3=-\dfrac{1}{9}\)
=>\(\left(3x-\dfrac{1}{2}\right)^3=-\dfrac{1}{27}\)
=>\(3x-\dfrac{1}{2}=-\dfrac{1}{3}\)
=>\(3x=\dfrac{1}{2}-\dfrac{1}{3}=\dfrac{1}{6}\)
=>\(x=\dfrac{1}{6}:3=\dfrac{1}{18}\)