Question

Given a+b+c=0 , abc ≠ 0 , a, b,c, are not equal to each other, find the value of

\dfrac{a^2}{bc} +\dfrac{b^2}{ac} +\dfrac{c^2}{ab}

Collected in the board: a+b+c=0 problems

Steven Zheng posted 3 years ago

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

Steven Zheng posted 3 years ago

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