Is this value correct or should it be simplified?

Question

Given that $a\neq p$, $b\neq q$, $c\neq r$, and $\begin{vmatrix} p & b & c \\ a & q & c \\ a & b & r \end{vmatrix} =0$ Then find the value of $\frac{p}{p-a}+\frac{q}{q-b}+\frac{r}{r-c} $

I got the solution as $$\frac{p}{p-a}+\frac{q}{q-b}+\frac{r}{r-c} = - 2\left[ \frac{a}{p-a}+\frac{b}{q-b}+\frac{c}{r-c} \right]$$

Should it be simplified or is there another solution to it that is more suitable?

Answer 1

Hint: Collect terms that have a common denominator.

You have $X=2Y$. Can you see another connection between $X$ and $Y$?

Answer 2

I got it figured out as below - $$\frac{p}{p-a}+\frac{q}{q-b}+\frac{r}{r-c} - \left[ \frac{a}{p-a}+\frac{b}{q-b}+\frac{c}{r-c} \right]= -3\left[ \frac{a}{p-a}+\frac{b}{q-b}+\frac{c}{r-c} \right]$$ $$\frac{p-a}{p-a}+\frac{q-b}{q-b}+\frac{r-c}{r-c}= -3 \left[ \frac{a}{p-a}+\frac{b}{q-b}+\frac{c}{r-c} \right]$$ $$3 = -3 \frac{a}{p-a}+\frac{b}{q-b}+\frac{c}{r-c} = $$

$$-1 = \frac{a}{p-a}+\frac{b}{q-b}+\frac{c}{r-c} = $$

So, $$\frac{p}{p-a}+\frac{q}{q-b}+\frac{r}{r-c} = -2\left[ -1 \right] = 2$$