If x^3+3ax^2+bx+c is a perfect cube .prove that b^3=27c^2
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!
To prove that b^3 = 27c^2 when the cubic equation x^3 + 3ax^2 + bx + c is a perfect cube, we need to show that the given equation can be expressed as (mx + n)^3 for some values of m and n.
Let's begin by assuming that the given equation can be expressed as (mx + n)^3. Expanding this, we get:
(mx + n)^3 = (mx)^3 + 3(mx)^2n + 3m(nx)^2 + n^3
= m^3x^3 + 3m^2nx^2 + 3mn^2x + n^3
Now, we can compare the coefficients of the two equations:
Coefficient of x^3: m^3 = 1 (as we are assuming the given equation is a perfect cube, and the coefficient of x^3 is 1)
Coefficient of x^2: 3m^2n = 3a (from the given equation, the coefficient of x^2 is 3a)
Coefficient of x: 3mn^2 = b (from the given equation, the coefficient of x is b)
Constant term: n^3 = c (from the given equation, the constant term is c)
From the coefficient of x^3, we can conclude that m = 1.
Substituting this value of m in the coefficient of x^2, we get: 3n = 3a, which simplifies to n = a.
Similarly, substituting the value of m in the coefficient of x, we get: 3n^2 = b, which simplifies to 3a^2 = b.
Finally, from the constant term, we have n^3 = c, which can be written as a^3 = c.
Now, let's calculate b^3 and 27c^2:
b^3 = (3a^2)^3 = 27a^6
27c^2 = 27(a^3)^2 = 27a^6
As we can see, b^3 and 27c^2 are equal, which proves that b^3 = 27c^2 when the given cubic equation x^3 + 3ax^2 + bx + c is a perfect cube.
In summary, we have shown that if the given cubic equation can be expressed as (mx + n)^3, with m = 1 and n = a, then b^3 = 27c^2.