\(\dfrac{1}{3}x+y+1=0\)
=>\(\dfrac{1}{3}x+y=-1\)
\(M=x^3+9x^2y+27xy^2+27y^3+27\)
\(=\left(x^3+9x^2y+27xy^2+27y^3\right)+27\)
\(=\left(x+3y\right)^3+27\)
\(=\left[3\left(x+\dfrac{1}{3}y\right)\right]^3+27\)
\(=27\left(x+\dfrac{1}{3}y\right)^3+27\)
\(=27\left(-1\right)^3+27=0\)