Question

Solve:

y^3 + 6y^2 = 27y

Answer 1

y(y^2+6y-27)=0
y(y^2 +9y -3y -27)=0
y { y(y+9) -3(y+9)}=0
y{ (y+9)(y-3)}=0
y(y+9)(y-3)=0
hence y=0,-9,3

Answer 2

y^3 +6y^2-27y=0
y(y-3)(y+9)=0
y=-9
y=0
y=3

Answer 3

=> y^3 + 6y^2 - 27Y = 0
=> y(Y^2 + 6y - 27) = 0
=> y [ y^2 + 9y -3y - 27) = 0
=> y[ y(y+9) -3(y+9) = 0
=> y (y-3) (y+9) = 0

Answer 4

the product can be zero if either of the factors r zero
y = 0 => y=0
y-3 = 0 => y = 3
y+9 = 0 => y = -9

so y = 0, 3, -9