Question
Answer
\dfrac{a^2}{bc} +\dfrac{b^2}{ac} +\dfrac{c^2}{ab} =\dfrac{a^3+b^3+c^3}{abc}
\because a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^3-ab-bc-ac)
\quad and
a+b+c=0
\therefore a^3+b^3+c^3-3abc=0
\dfrac{a^3+b^3+c^3}{abc} =3
\dfrac{a^2}{bc} +\dfrac{b^2}{ac} +\dfrac{c^2}{ab} =\dfrac{a^3+b^3+c^3}{abc}
\because a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^3-ab-bc-ac)
\quad and
a+b+c=0
\therefore a^3+b^3+c^3-3abc=0
\dfrac{a^3+b^3+c^3}{abc} =3