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Question

If α,β,γ are the roots of x3+px2+qx+r=0, then (β+r3α)(γ+α3β)(α+β3γ)=

A
3p3+16pq+64r
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B
3p316pq+64r
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C
3p316pq
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D
3p3+16pq
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Solution

The correct option is A 3p316pq+64r
As α,β,γ are the roots of x3+px2+qx+r=0
Gives
s1=α+β+γ=ps2=αβ+βγ+γα=qs3=αβγ=r
Now α+β+γ=pα+β=pγβ+γ=pαα+γ=pβ

Using this we get
(β+γ3α)(γ+α3β)(α+β3γ)=(p+4α)(p+4β)(p+4γ)=(p3+4p2(α+β+γ))+16p(αβ+βγ+γα)+64(αβγ)=(p3+4p2(p))+16p(q)+64(r)=3p316pq+64r

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