丁卯 show : a/(1 - a²) + b/(1 - b²) + c/(1 - c²) = 4abc/[(1 - a²)(1 - b²)(1 - c²)].?

Question

all a,b,c are not 士 1.

a + b + c = abc.

show : a/(1 - a²) + b/(1 - b²) + c/(1 - c²) = 4abc/[(1 - a²)(1 - b²)(1 - c²)].

Answer 1

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LHS = { a(1−b²)(1−c²) + b(1−a²)(1−c²) + c(1−a²)(1−b²) } / {(1−a²)(1−b²)(1−c²)}

Numerator expands to abc∑ab−∑a²(b+c)+∑a

Set u=∑a, v=∑ab, w=abc and noting that ∑a²(b+c) = ∑{a(ab+ac+bc)−abc} = uv−3w

Numerator = wv−(uv−3w)+u = (w−u)(v−1)+4w = 4w since w=u

∴ LHS = 4abc/{(1−a²)(1−b²)(1−c²)}

• aya2 years agoReport

  It took me 40minutes.
  I understood...but...from where comes this transformation.!
  I want to go where no math exists.