Question

If pq+qr+rp=0 then find the value of p^2/p^2-qr +q^2/q^2-rp + r^2/r^2-pq

Answer 1

Answer:

The value of

$\bf\frac{p^2}{p^2-qr}+\frac{q^2}{q^2-rp}+\frac{r^2}{r^2-pq}$       is 1

Step-by-step explanation:

Given:

$pq+qr+rp=0$.......(1)

Now,

$\frac{p^2}{p^2-qr}+\frac{q^2}{q^2-rp}+\frac{r^2}{r^2-pq}$

$=\frac{p^2}{p^2+pq+rp}+\frac{q^2}{q^2+pq+qr}+\frac{r^2}{r^2+qr+rp}$       (Using (1))

$=\frac{p^2}{p(p+q+r)}+\frac{q^2}{q(p+q+r)}+\frac{r^2}{r(p+q+r)}$

$=\frac{p}{p+q+r}+\frac{q}{p+q+r}+\frac{r}{p+q+r}$

$=\frac{p+q+r}{p+q+r}$

$=1$

$\implies\boxed{\bf\frac{p^2}{p^2-qr}+\frac{q^2}{q^2-rp}+\frac{r^2}{r^2-pq}=1}$