Ta có: $a+b+c=0$
$\Rightarrow (a+b+c)(a^2+b^2+c^2-ab-bc-ca)=0$
$\Leftrightarrow a^3+b^3+c^3=3abc$
$B=\dfrac{a^2}{a^2-b^2-c^2}+\dfrac{b^2}{b^2-c^2-a^2}+\dfrac{c^2}{c^2-a^2-b^2}$
$=\dfrac{a^2}{(b+c)^2-b^2-c^2}+\dfrac{b^2}{(c+a)^2-c^2-a^2}+\dfrac{c^2}{(a+b)^2-a^2-b^2}$
$=\dfrac{a^2}{2bc}+\dfrac{b^2}{2ca}+\dfrac{c^2}{2ab}$
$=\dfrac{a^3+b^3+c^3}{2abc}$
$=\dfrac{3abc}{2abc}=\dfrac{3}{2}$