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Question

# If the zeroes of the polynomial p(x)=ax3+3bx2+3cx+d are in AP, then 2b3−3abc+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 C 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

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