Let p,q and r be the distinct roots of the polynomial x^3 −22x^2 +80x−67. There exist real number A,B and C such that (1/(s^3 −22s^2 +80s−67)) = (A/(s−p)) + (B/(s−q)) + (C/(s−r)) for all real numbers s with s ∉ {p,q,r}.What is (1/A) + (1/B) + (1/C) ? (a) 243 (b) 244 (c) 245 (d)246 (e) 247


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Question Number 135566 by bemath last updated on 14/Mar/21
$Let p,q and r be the distinct roots of the polynomial x^3 −22x^2 +80x−67. There exist real number A,B and C such that (1/(s^3 −22s^2 +80s−67)) = (A/(s−p)) + (B/(s−q)) + (C/(s−r)) for all real numbers s with s ∉ {p,q,r}.What is (1/A) + (1/B) + (1/C) ? (a) 243 (b) 244 (c) 245 (d)246 (e) 247$
$L e t p, q a n d r b e t h e d i s t i n c t r o o t s$ $o f t h e p o l y n o m i a l x^{3} - 22 x^{2} + 80 x - 67 .$ $T h e r e e x i s t r e a l n u m b e r A, B a n d$ $C s u c h t h a t \frac{1}{s^{3} - 22 s^{2} + 80 s - 67} =$ $\frac{A}{s - p} + \frac{B}{s - q} + \frac{C}{s - r} f o r a l l r e a l n u m b e r s$ $s w i t h s \notin {p, q, r} . W h a t i s$ $\frac{1}{A} + \frac{1}{B} + \frac{1}{C} ?$ $(a) 243 (b) 244 (c) 245 (d) 246$ $(e) 247$
Commented by EDWIN88 last updated on 14/Mar/21

$very nice$
	Answered by EDWIN88 last updated on 14/Mar/21
	$⇔ 1 = A(s−q)((s−r)+B(s−p)(s−r)+C(s−p)(s−q) (•)s=p ⇒1=A(p−q)(p−r) ; (1/A)=(p−q)(p−r) similarly ⇒ { (((1/B)=(q−p)(q−r))),(((1/C)=(r−p)(r−q))) :} (1/A)+(1/B)+(1/C) = p^2 +q^2 +r^2 −pq−pr−qr = (p+q+r)^2 −3(pq+pr+qr) = 22^2 −3(80)= 244$
	$\Leftrightarrow 1 = A (s - q) ((s - r) + B (s - p) (s - r) + C (s - p) (s - q)$ $(∙) s = p \Rightarrow 1 = A (p - q) (p - r); \frac{1}{A} = (p - q) (p - r)$ $similarly \Rightarrow {\begin{cases} \frac{1}{B} = (q - p) (q - r) \\ \frac{1}{C} = (r - p) (r - q) \end{cases}$ $\frac{1}{A} + \frac{1}{B} + \frac{1}{C} = p^{2} + q^{2} + r^{2} - pq - pr - qr$ $= {(p + q + r)}^{2} - 3 (pq + pr + qr)$ $= 22^{2} - 3 (80) = 244$
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