using quadratic formula we get

first root as = (-b+sqrt(b^2-4ac))/2a

Second root =(-b-sqrt(b^2-4ac))/2a 

We have that one root is square of other so we get

(-b-sqrt(b^2-4ac))/2a = ((-b+sqrt(b^2-4ac))/2a)^2

=> -b-sqrt(b^2-4ac) = (b^2+b^2-4ac-2b*sqrt(b^2-4ac))/2a

=> -2ab-2a*sqrt(b^2-4ac) = 2b^2-4ac-2b*sqrt(b^2-4ac)

=> -ab-a*sqrt(b^2-4ac) = b^2-2ac-b*sqrt(b^2-4ac)

=> b^2-2ac+ab = sqrt(b^2-4ac)*(b-a)

squaring we get

b^4+4a^2c^2+a^2b^2-4b^2ac-4a^2bc+2b^3a=(b^2-4ac)*(b^2-2ab+a^2)

on simplification and dividing by a we get

b^3+a^2c+ac^2=3abc.