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Let's consider the expressions in the denominators. Since $a+b+c=0,$ $$a^2-bc=(-b-c{)}^2-bc=b^2+bc+c^2=b^2+c(b+c)=b^2-ac.$$Similarly, we can prove that $b^2-ac=c^2-ab.$

Let $x=a^2-bc=b^2-ac=c^2-ab.$ Then the sum is $$\frac{a^2{b}^2+a^2{c}^2+b^2{c}^2}{x^2}.$$Note that $$\begin{array}{rl}{x}^2& =(a^2-bc)(b^2-ac)\\ & =a^2{b}^2-a^{3}c-b^{3}c+ab{c}^2\\ & =a^2{b}^2-(a^3+b^3)c+ab{c}^2\\ & =a^2{b}^2-(a+b)(a^2-ab+b^2)c+ab{c}^2\\ & =a^2{b}^2+(a^2-ab+b^2)c^2+ab{c}^2\\ & =a^2{b}^2+a^2{c}^2+b^2{c}^2.\end{array}$$Therefore, $$\frac{a^2{b}^2+a^2{c}^2+b^2{c}^2}{x^2}=1.$$Thus, the given expression can only be equal to $\boxed{1}.$