Giải thích các bước giải:

Ta có:

$a+b+c=0$

$\to a+b=-c, b+c=-a, c+a=-b$

$\to \dfrac{a^2+b^2}{a+b}+\dfrac{b^2+c^2}{b+c}+\dfrac{c^2+a^2}{c+a}= \dfrac{a^2+b^2}{-c}+\dfrac{b^2+c^2}{-a}+\dfrac{c^2+a^2}{-b}$

$\to \dfrac{a^2+b^2}{a+b}+\dfrac{b^2+c^2}{b+c}+\dfrac{c^2+a^2}{c+a}=\dfrac{ab(a^2+b^2)+bc(b^2+c^2)+ca(c^2+a^2)}{-abc}$

$\to \dfrac{a^2+b^2}{a+b}+\dfrac{b^2+c^2}{b+c}+\dfrac{c^2+a^2}{c+a}=\dfrac{a^3b+ab^3+b^3c+bc^3+c^3a+ca^3}{-abc}$

$\to \dfrac{a^2+b^2}{a+b}+\dfrac{b^2+c^2}{b+c}+\dfrac{c^2+a^2}{c+a}=\dfrac{a^3(b+c)+b^3(c+a)+c^3(a+b)}{-abc}$

$\to \dfrac{a^2+b^2}{a+b}+\dfrac{b^2+c^2}{b+c}+\dfrac{c^2+a^2}{c+a}=\dfrac{a^3(-a)+b^3(-b)+c^3(-c)}{-abc}$

$\to \dfrac{a^2+b^2}{a+b}+\dfrac{b^2+c^2}{b+c}+\dfrac{c^2+a^2}{c+a}=\dfrac{-a^4-b^4-c^4}{-abc}$

$\to \dfrac{a^2+b^2}{a+b}+\dfrac{b^2+c^2}{b+c}+\dfrac{c^2+a^2}{c+a}=\dfrac{a^4+b^4+c^4}{abc}$

$\to \dfrac{a^2+b^2}{a+b}+\dfrac{b^2+c^2}{b+c}+\dfrac{c^2+a^2}{c+a}=\dfrac{a^3}{bc}+\dfrac{b^3}{ca}+\dfrac{c^3}{ab}$