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Question

If the zeroes of the polynomial p(x)=ax3+3bx2+3cx+d are in AP, then 2b33abc+a2d is equal to


A

3

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B

-3

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C

0

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D

Can't be determined

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Solution

The correct option is B

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=3caba×(2×ba)+db=3ca[ α+γ=2β=2×ba]
2b2a2+db=3ca 2b3+da2=3abc 2b33abc+a2d=0


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