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Let a, b, and c, be nonzero real numbers such that a+b+c=0. Compute the value of 

 Apr 5, 2020
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Hello SpongeBobRules24!


Distribute:

(1) \(\frac{a}{b}+\frac{a}{c}+\frac{b}{a}+\frac{b}{c}+\frac{c}{a}+\frac{c}{b}\)

 

Combine like terms (fractions with the same denominator):

(1) \(\frac{b+c}{a}+\frac{a+c}{b}+\frac{a+b}{c}\)

 

Consider:

It was given that 

(2) \(a+b+c=0\)

 

We solve for the numerators of the three fractions in (1).

Solving:

Let us do the fraction with \(a\) as the denominator.

\(b+c=-a\)

We substitute this into (1):

\(\frac{-a}{a}=-1\)

 

Our equation is now:

\(-1+\frac{a+c}{b}+\frac{a+b}{c}\)


We repeat the process of solving for the numerators of the fractions in (1) and eventually get:
\(-1+-1+-1\)

 

Which evaluates to \(\boxed{-3}\)

.
 Apr 5, 2020
edited by AnExtremelyLongName  Apr 5, 2020

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