We'll start by manipulating the given expression \(\frac{a^2}{bc} + \frac{b^2}{ac} + \frac{c^2}{ab}\).
Given that \(a + c = -b\), we can rewrite this as \(b = -a - c\).
Now, let's substitute \(b\) in the expression:
\[
\frac{a^2}{bc} + \frac{b^2}{ac} + \frac{c^2}{ab} = \frac{a^2}{c(-a - c)} + \frac{(-a - c)^2}{ac} + \frac{c^2}{a(-a - c)}
\]
Simplify further:
\frac{-a^2}{ac + c^2} + \frac{a^2 + 2ac + c^2}{ac} - \frac{c^2}{a^2 + ac}
= \frac{-a^2}{ac + c^2} + \frac{a^2 + 2ac + c^2}{ac} - \frac{c^2}{a(a + c)}
= \frac{-a^2}{ac + c^2} + \frac{a^2 + 2ac + c^2}{ac} - \frac{c^2}{a(-b)}
= \frac{-a^2}{ac + c^2} + \frac{a^2 + 2ac + c^2}{ac} + \frac{c^2}{ab}
= \frac{-a^2 + a^2 + 2ac + c^2 + c^2}{ac + c^2}
= \frac{2ac + 2c^2}{ac + c^2}
= \frac{2c(a + c)}{c(a + c)}
= 2
So, if \(a + c = -b\), then \(\frac{a^2}{bc} + \frac{b^2}{ac} + \frac{c^2}{ab} = 2\).
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Answers & Comments
We'll start by manipulating the given expression \(\frac{a^2}{bc} + \frac{b^2}{ac} + \frac{c^2}{ab}\).
Given that \(a + c = -b\), we can rewrite this as \(b = -a - c\).
Now, let's substitute \(b\) in the expression:
\[
\frac{a^2}{bc} + \frac{b^2}{ac} + \frac{c^2}{ab} = \frac{a^2}{c(-a - c)} + \frac{(-a - c)^2}{ac} + \frac{c^2}{a(-a - c)}
\]
Simplify further:
\[
\frac{-a^2}{ac + c^2} + \frac{a^2 + 2ac + c^2}{ac} - \frac{c^2}{a^2 + ac}
\]
\[
= \frac{-a^2}{ac + c^2} + \frac{a^2 + 2ac + c^2}{ac} - \frac{c^2}{a(a + c)}
\]
\[
= \frac{-a^2}{ac + c^2} + \frac{a^2 + 2ac + c^2}{ac} - \frac{c^2}{a(-b)}
\]
\[
= \frac{-a^2}{ac + c^2} + \frac{a^2 + 2ac + c^2}{ac} + \frac{c^2}{ab}
\]
\[
= \frac{-a^2 + a^2 + 2ac + c^2 + c^2}{ac + c^2}
\]
\[
= \frac{2ac + 2c^2}{ac + c^2}
\]
\[
= \frac{2c(a + c)}{c(a + c)}
\]
\[
= 2
\]
So, if \(a + c = -b\), then \(\frac{a^2}{bc} + \frac{b^2}{ac} + \frac{c^2}{ab} = 2\).