Question

Разложите на множетели 1/9a^3+3
x^6 - y^6 y^5-2y^3+y

Answer 1

$\frac{1}{9} a^{3} +3=3( \frac{1}{27} a^{3} +1)=3( \frac{1}{3}a+1)( \frac{1}{9}a^{2} -\frac{1}{3}a+1)= \\ \\ = 3\frac{(a+3)( a^{2}-3a+9 )}{27} =\frac{(a+3)( a^{2}-3a+9 )}{9}$


$x^{6} - y^{6} = (x+y)(x-y)(x^{4}+ x^{2} y^{2}+ y^{4} )$


$y^{5} -2 y^{3} +y=y( y^{4} -2 y^{2} +1)=y( y+1)^{2} (y-1)^{2}$