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:
$$\sum_{i=1}^{3}\frac{1}{x^2_i} = \left({\sum_{i=1}^{3}\frac{1}{x_i}}\right)^2.$$

Hint:
What is value of  $\frac{1}{x_1.x_2}+\frac{1}{x_2.x_3}+\frac{1}{x_3.x_1}$ ,when $x_1+x_2+x_3=0$.
If you got the value of $\frac{1}{x_1.x_2}+\frac{1}{x_2.x_3}+\frac{1}{x_3.x_1}$, then proceed by expanding $(\sum_{i=1}^3 \frac{1}{x_i})^2$ by using the formula $(a+b+c)^2= a^2+b^2+c^2+ 2(ab+bc+ac)$
