If the zeroes of the polynomial p(x)=ax3+3bx2+3cx+d are in AP, then 2b3−3abc+a2d is equal to
0
Let the zeroes be α, β and γ
Then, α+β+γ=−3ba . . . (i)
αβ+βγ+γα=3ca . . . (ii)
and αβγ=−da . . . (iii)
Now, α, β and γ are in AP.
⇒ 2β=α+γ
From Eq. (i), 3β=−3ba
⇒ β=−ba
Also, from Eq. (iii),
αβγ=−da⇒ (αγ)(−ba)=−da⇒ αγ=db
From Eq. (ii),
β(α+γ)+αγ=3ca⇒−ba×(2×−ba)+db=3ca[∵ α+γ=2β=2×−ba]
⇒ 2b2a2+db=3ca⇒ 2b3+da2=3abc⇒ 2b3−3abc+a2d=0