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Question

If α,β are the roots of ax2+bx+c=0 then (aα+b)3 +(aβ+b)3


A

a3-2abc

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B

b3 – 3abc

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C

(c3 – 3abc)/ b3c3

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D

(b3 – 3abc)/ a3c3

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Solution

The correct option is D

(b3 – 3abc)/ a3c3


α+β= ba , αβ=ca

α ≠ 0 is root of the equation ax2+bx+c =0 then aα2+bα+c=0 => aα+b=ca

similarly αβ+b = cβ

so (aα+b)3 +(aβ+b)3 = (α3+β3)c3 = [(α+β)33αβ(α+β)]c3 = b33abca3c3


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