Question : If $p+q+r=0$, then what is the simplified value of the expression $(\frac{p^2}{p^2-q r}+\frac{q^2}{q^2-p r}+\frac{r^2}{r^2-p q})$?
Option 1: 0
Option 2: 2
Option 3: 1
Option 4: – 1
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Correct Answer: 2
Solution : Given: $p+q+r=0$ To find: $(\frac{p^2}{p^2-q r}+\frac{q^2}{q^2-p r}+\frac{r^2}{r^2-p q})$ By putting the values p = 1, q = –1, and r = 0, the equation $p+q+r=0$ satisfies. $(\frac{p^2}{p^2-q r}+\frac{q^2}{q^2-p r}+\frac{r^2}{r^2-p q})$ = $(\frac{1^2}{1^2-(-1)× 0}+\frac{(-1)^2}{(-1)^2-1×0}+\frac{0^2}{0^2-1×(-1)})$ = 1 + 1 + 0 = 2 Hence, the correct answer is 2.
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