djb86 wrote: Given that \[\frac{a-b}{c-d}=2\quad\text{and}\quad\frac{a-c}{b-d}=3\] for certain real numbers $a,b,c,d$, determine the value of \[\frac{a-d}{b-c}.\] $\frac{a-d}{b-c}=\frac{\frac{a-b}{c-d}\frac{a-c}{b-d}-1}{\frac{a-b}{c-d}-\frac{a-c}{b-d}}=-5$

$a=4c-3b,d=2b-c,$ $\frac{a-d}{b-c}=\frac{5c-5b}{b-c}=-5.$

$a= b +2c -2d$ $a = c +3b -3d$ $b+2c - 2d = c + 3b -3d$ $2b - c -d=0$ $b-c = d-b = \frac{c-a}{3}$ $a-b = 2c - 2d$ $a-c = 3b -3d$ $2c = a -b +2d$ $c= a-3b +3d$ $2c = 2a - 6b + 6d$ $0 = a-5b +4d$ $a = 5b-4d$ $a-d - 5(b-d)$ $= 5 \cdot \frac{(a-c)}{3}$ $\frac{a-d}{b-c}= \frac{5}{3} \cdot \frac{(a-c)}{ \frac{(c-a)}{3}}$ $ = -5$.

Let $a, b$ and $c$ be nonzero and distinct complex numbers. Prove that $$\frac{a^2(1+b^2c^2)}{(a-b)(a-c)}+\frac{b^2(1+c^2a^2)}{(b-c)(b-a)}+\frac{c^2(1+a^2b^2)}{(c-a)(c-b)} $$is integer and determine its value. (Juan Jos'e Egozcue and Jos'e Luis D'ıaz Barrero)

sqing wrote: Let $a, b$ and $c$ be nonzero and distinct complex numbers. Prove that $$\frac{a^2(1+b^2c^2)}{(a-b)(a-c)}+\frac{b^2(1+c^2a^2)}{(b-c)(b-a)}+\frac{c^2(1+a^2b^2)}{(c-a)(c-b)} $$is integer and determine its value. (Juan Jos'e Egozcue and Jos'e Luis D'ıaz Barrero) We rewrite it as $$\frac{a^2(1+b^2c^2)(b-c)+b^2(1+c^2a^2)(c-a)+c^2(1+a^2b^2)(a-b)}{(a-b)(a-c)(b-c)}$$which after some cancellation becomes $$\frac{a^2(b-c)+b^2(c-a)+c^2(a-b)}{(a-b)(a-c)(b-c)}=\frac{a^2b-a^2c+b^2c-b^2a+c^2a-c^2b}{a^2b-a^2c+b^2c-b^2a+c^2a-c^2b}=1.$$