If the ratios of the polynomial ax^{3}+3bx^{2}+3cx+d are in AP, Prove that 2b^{3}-3abc+a^{2}d=0
Ans: Let p(x) = ax^{3} + 3bx^{2} + 3cx + d and α , β , r are their three Zeros but zero are in AP
let α = m - n , β = m, r = m + n
sum = α+β+ r = - b/a
substitute this sum , to get = m= -b/a
Now taking two zeros as sum αβ +β r +αr = c a
(m-n)m + m(m+n) + (m + n)(m - n) = 3c/a
Solve this problem , then we get
3b^{2} - 3ac/a^{2}^{ }= n^{2}
Product αβ r = d/ a
(m-n)m (m+n) = -d/a
(m^{2} -n^{2})m = - d/a
[(-b/a)^{2}-(3b^{2} -3ac/a^{2})](-b/a) = -d/a
Simplifying we get
2b^{3} - 3abc + a^{2} d = 0