If x^3+3ax^2+bx+c is a perfect cube .prove that b^3=27c^2

Answer 1

to have your expanded cubic start with x^2
you must have cubed a binomial of the form (x + m)^3

if we expand (x+m)^3 we get
x^3 + 3x^2 m + 3x m^2 + m^3
compare that with your given
x^3 + 3ax^2 + bx + c
we can say:
3a = 3 ---> a = 1

3xm^2 = bx ---> b = 3m^2
and c = m^3

so we want to prove b^3 = 27c^2

LS = b^3 = 27m^6
RS = 27(m^6) = LS

all done!