If α,β,γ are the roots of the equation x3−6x2+11x−6=0 and if a=α2+β2+γ2, b=αβ+βγ+γα and c=(α+β)(β+γ)(γ+α), then the correct inequality among the following is
Given equation x3−6x2+11x−6=0 has the roots α,βγ Given, ‌‌a=α2+β2+γ2 . . . (i) ‌b=αβ+βγ+γα . . . (ii) ‌c=(α+β)(β+γ)(γ+α) . . . (iii) In cubic equation the sum of the roots α+β+γ=−(‌
−6
1
)=6 αβ+βγ+γα=(‌
11
1
)=11 product of the roots α⋅β⋅γ=−(‌
−6
1
)=6 From Eq. (ii), b=11 From Eq. (i), a=α2+β2+γ2 ⇒a=(α+β+γ)2−2(αβ+βγ+γα) ⇒a=(6)2−2(11)⇒36−22 ⇒a=14 From Eq. (iii) ‌c=(α+β)(β+γ)(γ+α) ‌=(αβ+β2+αγ+βγ)(γ+α) ‌=αβγ+β2γ+αγ2+βγ2+α2β ‌+αβ2+α2γ+αβγ ‌⇒c=[(α+β+γ)−γ][(α+β+γ)−α] ‌[(α+β+γ)−β] ‌=(6−γ)(6−α)(6−β) ‌=(36−6γ−6α+αγ)(6−β) ‌=216−36γ−36α+6αγ−36β+6γβ ‌+6αβ−αβγ ‌=216−αβγ+6(αβ+βγ+γα)−36(α+β+γ) ‌=216−6+6(11)−36(6) ‌=210+66−216=60 ‌‌ Hence, ‌‌‌c=60 ‌⇒‌‌b<a<c Alternate method Given equation is x3−6x2+11x−6=0 ⇒‌‌(x−1)(x−2)(x−3)=0 The roots of this equation are 1,2 , and 3 . Let α=1,β=2γ=3 Now, ‌‌a=α2+β2+γ2 ‌=1+4+9=14 b‌=αβ+βγ+γα ‌=2+6+3=11 c‌=(α+β)(β+γ)(γ+α) ‌=3⋅5⋅4=60 From the values of a,b and c it is clear that b<a<c.