Given,

$${1 \over {x + p}} + {1 \over {x + q}} = {1 \over r}$$

$$ \Rightarrow $$$$\,\,\,$$ $${{x + p + x + q} \over {\left( {x + p} \right)\left( {x + q} \right)}} = {1 \over r}$$

$$ \Rightarrow $$$$\,\,\,$$ (2x + p + q) r = x² + px + qx + pq

$$ \Rightarrow $$$$\,\,\,$$ x² + (p + q $$-$$ 2r) x + pq $$-$$ pr $$-$$ qr = 0

Let $$\alpha $$ and $$\beta $$ are the roots,

$$\therefore\,\,\,$$ $$\alpha $$ + $$\beta $$ = $$-$$ (p + q $$-$$ 2r)

and $$\alpha $$ $$\beta $$ = pq $$-$$ pr $$-$$ qr

Given that, $$\alpha $$ = $$-$$ $$\beta $$ $$ \Rightarrow $$ $$\alpha $$ + $$\beta $$ = 0

$$\therefore\,\,\,$$ $$-$$ (p + q $$-$$ 2r) = 0

Now, $$\alpha $$² + $$\beta $$²

= ($$\alpha $$ + $$\beta $$)² $$-$$ 2$$\alpha $$ $$\beta $$

= ($$-$$ (p + q $$-$$ 2r))² $$-$$ 2 (pq $$-$$ pr $$-$$ qr)

= p² +q² + 4r² + 2pq $$-$$ 4pr $$-$$ 4qr $$-$$ 2pq + 2pr + 2qr

= p² + q² + 4r² $$-$$ 2pr $$-$$ 2qr

= p² + q² $$-$$ 2r (p + q $$-$$ 2r)

= p² + q² $$-$$ 2r (0)

= p² + q²