<s>[INST] Evaluating $\frac{1}{a^2-ac-ab+bc} + \frac{2}{b^2-ab-bc+ac} + \frac{1}{c^2-ac-bc+ab}$ for $a=1.69$, $b=1.73$, and $c=0.48$ An interesting question on algebra:

If $a=1.69$, $b=1.73$, and $c=0.48$, find the value of$\\$
$$\frac{1}{a^2-ac-ab+bc} + \frac{2}{b^2-ab-bc+ac} + \frac{1}{c^2-ac-bc+ab}$$

By substituting the values, I got the answer, but I want to know if there is any short algebraic way because putting the values was a lot time consuming.
Thanks.
 [/INST] The expression $L$ is equal to
\begin{align}
L &= \frac{1}{(a-b)(a-c)}+\frac{2}{(b-c)(b-a)}+\frac{1}{(c-a)(c-b)} \\
&= \frac{-(b-c)-2(c-a)-(a-b)}{(a-b)(b-c)(c-a)} \\
&= \frac{a-c}{(a-b)(b-c)(c-a)} \\
&=- \frac{1}{(a-b)(b-c)} \\
\end{align}
and it's easy now to calculate $L = 20$.
</s>