Prove that 2b3-3abc+a2d=0, Mathematics

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

Posted Date: 4/8/2013 2:14:38 AM | Location : United States

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