[tex]\frac{2}{3}(1-y)(1+y)-\frac{2}{5}(y-3)(-y)+\frac{1}{15}(y-2)(y+2)=\\
\\
folosim\:(a-b)(a+b)=a^2-b^2\\
\\
=\frac{2}{3}(1-y^2)-\frac{2}{5}(-y^2+3y)+\frac{1}{15}(y^2-)=\\
\\
aducem\:la\:acelasi\:numitor,\:care\:este\:15\\
\\
=\frac{10(1-y^2)-6(-y^2+3y)+y^2-4}{15}=\frac{10-10y^2+6y^2-18y+y^2-4}{15}=\\
\\
=\frac{-3y^2-18y+6}{15}=\frac{-y^2-6y+2}{5}[/tex]
