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