[Ne[S]zA]

Let $\dfrac{a+b}{4}=\dfrac{b+c}{5}=\dfrac{c+a}{6}=k$
we have $a=\dfrac{5k}{2},b=\dfrac{3k}{2},c=\dfrac{7k}{2}$
therefore $\dfrac{ab+bc+ca}{a^2+b^2+c^2}=\dfrac{71}{83}$
____________________________________________________________________

Solve the equation
$$\sqrt[3]{2011x-2554} +\sqrt[3]{2011x+2554}=\sqrt[3]{4022x}$$