x^4-2x^3y+4/3x^2y^2-8/27xy^3
=x[x^3-2x^2y+4/3xy^2-(2/3y)^3]
=x{[x^3-(2/3y)^3]-2xy(x-2/3y)}
=x[(x-2/3y)(x^2+2/3xy+4/9y^2)-2xy(x-2/3y)]
=x(x-2/3y)(x^2+2/3xy+4/9y^2-2xy)
=x(x-2/3y)(x^2-4/3xy+4/9y^2)
=x(x-2/3y)(x-2/3y)^2
=x(x-2/3y)^3
解:原式=(x^2-2x-6/y^2-8+x^2/y-8)/27xy^3