Don't need hardwork

$E= \dfrac{500^2}{(bc)^2}+\dfrac{500^2}{(ac)^2}+\dfrac{500^2}{(ab)^2}$ .

$\implies E=(500)^2(\dfrac{a^2}{(abc)^2}+\dfrac{b^2}{(abc)^2}+\dfrac{c^2}{(abc)^2})$ .

Using $abc=\frac{500}{3}$ we get

$E=(500)^2(\dfrac{9 a^2}{(500)^2}+\dfrac{9 b^2}{(500))^2}+\dfrac{9 c^2}{(500)^2})$ .

$E=9(a^2+b^2+c^2)$

$E=9((a+b+c)^2-2(ab+bc+ca))$

Using $a+b+c=\dfrac{35}{3}$ and $ab+bc+ca=0$ we get

$E=9((\dfrac{35}{3})^2 - 2 \times 0)= (35)^2 = \boxed{1225}$

Using Vieta's Formula we have:- $a+b+c=\frac{35}{3} \\ ab+bc+ca=0 \\ abc=\frac{-500}{3}$

$\frac{500^{2}}{(bc)^{2}} = (\frac{500}{\frac{-500}{3a}})^{2}$

Similarly we will get a equation at last which is equal to:-

$9(a^{2}+b^{2}+c^{2}) \\ 9[(a+b+c)^{2}-2(ab+bc+ca)] \\ =9(\frac{35}{3})^{2} \\ =\boxed{1225}$

Another way is to determine the roots. As we solve the equation we get x=-10/3, 10, and 5 Just plug in those values