Question

If the zeroes of the polynomial f(x)= ax³ + 3bx²+ 3cx +d are in A.P.,prove that 2b³– 3abc + a² d=0

Answer 1

Let the roots be p,q and r
we know that
1) p.q.r= -d/a
2) p+ q+r = -3b/a
3) pq+rq+rp = 3c/a

Since the roots are in ap , we can assume the root to p-m , p and p+ m ( m is the common difference )
= p-m+p+p+m = -3b/a
= 3p = -3b/a
= p= -b/a

(p-m) (p) (p+m) = -d/a
p.(p^2-m^2) = -d/a
since p= -b/a , we get p^2-m^2 = d/b

apply third condition
p(p-m) + p( p+m) + (p-m)(p+m) =3c/a
2p^2 + ( p^2-m^2)= 3c/a

substituting for p and p^2-m^2
2b^2/a^2 +d/b = 3c/a

simplifying the gives ,
2b^3 + a^2d = 3abc
2b^3 -3abc +a^2d = 0

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