You can put this solution on YOUR website! 27y^3 +125=0,therefore
(3y)^3 +(5)^3 =0=>
(3y + 5)((3y)^2 -3y*5 +5^2)=0. Here I use the identity a^3 +b^3 = (a+b)(a^2 -ab +b^2).
Therefore 3y +5 =0 gives y= -5/3.The other factor is a quadratic equated to zero gives 9y^2 -15y +25 =0,therefore,
y= [-(-15) + or - sqrt( (-15)^2 -4*9*25)]/(2*9)
=[15 + or -15sqrt(225-900)]/18
=[15+ or -15sqrt(1-4))/18
=15[1 + or -sqrt(-3)]/18
=5[1+ or - sqrt(-3)]/6
(