The three roots of the cubic $ 30 x^3 – 50x^2 + 22x – 1$ are distinct real numbers strictly between $ 0$ and $ 1$. If the roots are $p$, $q$, and $r$, what is the sum
$ \frac{1}{1-p} + \frac{1}{1-q} +\frac{1}{1-r} ?$
Expanding $(x-p)(x-q)(x-r)=x^{3}-(r+p+q)x^{2}+(qr+pr+pq)-pqr.$
Then, $r+q+p=50,$ and $rqp=1.$
I'm now stumped.
Best Answer
Hint. Simply combining the three fractions: $$\begin{split}\frac1{1-p}+\frac1{1-q}+\frac1{1-r}&=\frac{(1-q)(1-r)+(1-p)(1-r)+(1-p)(1-q)}{(1-p)(1-q)(1-r)}\\&=\frac{3-2p-2q-2r+pr+qr+pq}{1-p-q-r+pq+qr+rp-pqr}\\ &=\frac{3-2(p+q+r)+(pr+qr+pq)}{1-(p+q+r)+(pq+qr+rp)-(pqr)}.\end{split}$$ Now, can you finish off by using Vieta's formulas?