9 September 2011

• curprev 16:0016:00, 9 September 2011 Negativebplusorminus talk contribs 651 bytes +651 Created page with "We see that, by Vieta, <math>a+b+c=1</math>. Thus, <cmath>\frac{a}{a+b+c-a}+\frac{b}{a+b+c-b}+\frac{c}{a+b+c-c}=\frac{a}{1-a}+\frac{b}{1-b}+\frac{c}{1-c}</cmath> which we can ma..."