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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
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