\(s^3 - 22s^2 + 80s - 67\) should be factorised as \((s-p)(s-q)(s-r)\), as \(p,q\) and \(r\) are the roots of the polynomial \(x^3 - 22x^2 + 80x - 67\).
So, \(1=A(s-q)(s-r)+B(s-r)(s-p)+C(s-p)(s-q)\)

Now for \(s=p\), we would get, \(\frac{1}{A}=(p-q)(p-r)\)
Similarly, for \(s=q\), we get, \(\frac{1}{B}=(q-p)(q-r)\)
And, for \(s=r\), we get, \(\frac{1}{C}=(r-p)(r-q)\)
\(\therefore\) Adding them we can get, \(\frac{1}{A}+\frac{1}{B}+\frac{1}{C}=p^2+q^2+r^2-pq-qr-rp\)
Now form the given equation we can obtain the sum of the roots,
\(p+q+r=22\)
Also the sum of the roots taken two at a time \(pq+pr+rq=80\)
So, \(p^2+q^2+r^2=(p+q+r)^2-2(pq+pr+rq)\)
\(\Rightarrow p^2+q^2+r^2=22^2-160\)
\(\Rightarrow p^2+q^2+r^2=484-160=324\)
So, \(\frac{1}{A}+\frac{1}{B}+\frac{1}{C}=324-80=\boxed{244}\)

• This reply was modified 1 year, 3 months ago by Deepan Dutta.