+0 \(a+b+c=1,\:\frac{1}{a+b-c}+\frac{1}{b+c-a}+\frac{1}{c+a-b}=1,\:abc=? \)
+195 We can rewrite the given equation as:
ab−c2a+b−c+bc−a2b+c−a+ca−b2c+a−b=1.
Notice that the left-hand side of the equation resembles the sum of the reciprocals of three numbers, but with some extra terms. Let's manipulate the equation to exploit this similarity.
First, we can factor the quadratic terms in the denominators:
(a−c)(b)a+b−c+(b−a)(c)b+c−a+(c−b)(a)c+a−b=1.
Then, we can multiply both sides of the equation by the common denominator, which is the product of all the denominators:
(a−c)(b)⋅(a−c)(b)a+b−c+(b−a)(c)⋅(b−a)(c)b+c−a+(c−b)(a)⋅(c−b)(a)c+a−b=(a−c)(b)⋅(b−a)(c).
Expanding both sides, we get:
a(b+c−a)+b(c+a−b)+c(a+b−c)=(a−c)(b)(c−a).
This simplifies to:
3abc=abc−a2b−ab2−a2c−ac2−b2c.
Since a+b+c=1, we can substitute:
3abc=abc−(a2+b2+c2)−(ab+ac+bc).
We know that (a+b+c)2=a2+b2+c2+2(ab+ac+bc), so:
3abc=abc−(a+b+c)2+2(ab+ac+bc).
Substituting a+b+c=1, we get:
3abc=abc−1+2(ab+ac+bc).
Finally, we can factor out abc to find:
2abc=abc−1.
Solving for abc, we get:
abc = 1/3.
Therefore, the answer is abc = 1/3.
