0,36х^4-25/9у^6 плзплз

Решите задачу:

$0,36y^4- \frac{25}{9} y^6=(0.6)^2*y^4- \frac{(5)^2}{(3)^2}*y^2*y^4 =y^4*[(0.6)^2- (\frac{5}{3})^2y^2 ]=$
$=y^4*[( \frac{3}{5} )^2- (\frac{5}{3})^2y^2 ]=y^4*[( \frac{3}{5} )^2*(\frac{5}{3} )^2*( \frac{3}{5} )^2- (\frac{5}{3})^2y^2 ]=$
$=(\frac{5}{3} )^2y^4*[( \frac{3}{5} )^2*( \frac{3}{5} )^2-y^2] =(\frac{5}{3} )^2y^4*[(( \frac{3}{5} )^2)^2-y^2]=$
$=(\frac{5}{3} )^2y^4*[(( \frac{3}{5} )^2)^2-(y)^2]=(\frac{5}{3} )^2y^4[( \frac{3}{5} )^2-y]*[( \frac{3}{5} )^2+y]=$
$=(\frac{5}{3} )^2y^4[ 0.36-y]*[0.36+y]$