If  the  ratios  of  the  polynomial ax³+3bx²+3cx+d  are  in  AP,  Prove  that  2b³-3abc+a²d=0

Ans: Let p(x) = ax³ + 3bx² + 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

3b²  - 3ac/a² = n²

Product αβ r = d/ a

(m-n)m (m+n) = -d/a

(m² -n²)m = - d/a

[(-b/a)²-(3b² -3ac/a²)](-b/a) = -d/a

Simplifying we get

2b³ - 3abc + a² d = 0