${\left(0.2x+\frac{2}{5}y\right)}^3-\left(0.2x+\frac{2}{5}y\right)(\frac{x^2}{25}+\frac{4}{25}{y}^2-\frac{2\mathrm{xy}}{25})-0.08(x(\frac{3\mathrm{xy}}{5}+y)+3y(\frac{2\mathrm{xy}}{5}-\frac{x}{3}))=$

${=\left(\frac{1}{5}x+\frac{2}{5}y\right)}^3-\left(\frac{1}{5}x+\frac{2}{5}y\right)(\frac{x^2}{25}+\frac{4}{25}{y}^2-\frac{2\mathrm{xy}}{25})-\frac{2}{25}(x(\frac{3\mathrm{xy}}{5}+y)+3y(\frac{2\mathrm{xy}}{5}-\frac{x}{3}))=$

$=\frac{1}{125}{x}^3+\frac{12}{125}x{y}^2+\frac{6}{125}{x}^{2}y+\frac{8}{125}{y}^3-(\frac{1}{125}{x}^3+\frac{8}{125}{y}^3)-\frac{2}{25}(\frac{3}{5}{x}^{2}y+\mathrm{xy}+\frac{6}{5}x{y}^2-\mathrm{xy})=$

$=\frac{1}{125}{x}^3+\frac{12}{125}x{y}^2+\frac{6}{125}{x}^{2}y+\frac{8}{125}{y}^3-\frac{1}{125}{x}^3-\frac{8}{125}{y}^3-\frac{6}{125}{x}^{2}y-\frac{2}{25}\mathrm{xy}-\frac{12}{125}x{y}^2+\frac{2}{25}\mathrm{xy}=$

$=(\frac{1}{125}-\frac{1}{125})x^3+(\frac{8}{125}-\frac{8}{125})y^3+(\frac{12}{125}-\frac{12}{125})x{y}^2+(\frac{6}{125}-\frac{6}{125})x^{2}y+(-\frac{2}{25}+\frac{2}{25})\mathrm{xy}=0$