For $x_1,x_2,x_3\in\mathbb R$ that $x_1+x_2+x_3=0$ show that $\sum_{i=1}^{3}\frac{1}{x^2_i} =({\sum_{i=1}^{3}\frac{1}{x_i}})^2$ Show that if $ x_1,x_2,x_3 \in \mathbb{R}$ , and $x_1+x_2+x_3=0$ , we can say that:
$$\sum_{i=1}^{3}\frac{1}{x^2_i} = \left({\sum_{i=1}^{3}\frac{1}{x_i}}\right)^2.$$

Take the equatin $x_1+x_2+x_3=0$, divide by $x_1x_2x_3$, multiply by $2$, add $x_1^{-2}+x_2^{-2}+x_3^{-2}$. This is essentially reverse-engineered from taking the suspected equality, multiplying out the right-hand side, subtracting out the left-hand side and multiplying by $x_1x_2x_3$...
