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Question

# If α,β,γ are roots of the equation x3+2x−3=0, then the equation whose roots are (α−β)(α−γ),(β−γ)(β−α),(γ−α)(γ−β) is:

A
6x3+x2+275=0
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B
x3+6x2+275=0
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C
x36x2275=0
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D
x3+6x2275=0
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Solution

## The correct option is D x3+6x2−275=0x3+2x−3=0(x−1)(x2+x+3)=0Hence,Let γ=1Now, α+β=−1 and α.β=3Sum of roots(α−β)(α−γ)+(β−γ)(β−α)+(γ−α)(γ−β)=(α−β)(α−γ+γ−β)+(1−α)(1−β)=(α−β)2+1−(α+β)+α.β=(α+β)2−4α.β+1−(α+β)+α.β=1−12+1−(−1)+3=1−12+2+3=−6Hence,Sum of roots is =−ba=−6ba=6Since, a=1, b=6 Now product of roots is (α−β)(α−γ)(β−γ)(β−α)(γ−α)(γ−β)=−(α−β)2(1−α)2(1−β)2=−(α−β)2[1−(α+β)+α.β]2=−[(α+β)2−4α.β][1−(α+β)+α.β]2=−[1−12][1+1+3]2=−(−11)(25)=275=−caHence,c=−275Hence, the required equation is x3+6x2−275=0.

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