2021/090) If $p+q+r=0$ then what is $\frac{(p+q)^2}{3pq}+\frac{(q+r)^2}{3qr} +\frac{(r+p)^2}{3rp}$

We have p+q + r = 0

So $p^3 + q^3 + r^3 = 3pqr\cdots(1)$

And  p+q = -r

or $(p+q)^2 = r^2\cdots(2)$

Similary $(q+r)^2 = p^2\cdots(3)$

$(r+p)^2 = q^2\cdots(4)$

Hence $\frac{(p+q)^2}{3pq}+\frac{(q+r)^2}{3qr} +\frac{(r+p)^2}{3rp}$

$= \frac{r^2}{3pq}+\frac{p^2}{3qr} +\frac{q^2}{3rp}$ (from (2), (3), (4)

$= \frac{r^3+q^3+ p^3}{3pqr} = \frac{3pqr}{3pqr} = 1$  (using (1))