Since α,β and γ are the roots of the equation x3−10x2+7x+8‌=0,‌ then ‌ α+β+γ‌=10 . . . (i) αβ+βγ+γα‌=7 . . . (ii) and αβγ‌=−8 . . . (iii) On squaring equation (i) both sides, we get ‌α2+β2+γ2+2(αβ+βγ+γα)=100 ‌⇒‌‌α2+β2+γ2=100−2‌ (7) ‌‌‌‌ [from Eqs (ii)] ‌ ‌⇒α2+β2+γ2=86 . . . (iv) ‌‌ Now, ‌‌‌‌
1
α
+‌
1
β
+‌
1
γ
‌=‌
βγ+αβ+γα
αβγ
=‌
7
−8
[from Eqs (ii) and (iii)] Again now, ‌‌‌
α
βγ
+‌
β
γα
+‌
γ
αβ
‌=‌
α2+β2+γ2
αβγ
=‌
86
−8
[from Eqs (iii) and (iv)] ‌=−‌
43
4
From the above discussion we see that option (c) is correct.