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Question Number 10067 by konen last updated on 22/Jan/17
((a×b×c)/(a+b))=−(1/2)  ((a×b×c)/(b+c))=(1/2)  ((a×b×c)/(a+c))=1  ⇒a×b×c=?
a×b×ca+b=12a×b×cb+c=12a×b×ca+c=1a×b×c=?
Answered by mrW1 last updated on 22/Jan/17
let u=a×b×c  ⇒a+b=−2u    ...(i)  ⇒b+c=2u     ...(ii)  ⇒a+c=u      ...(iii)  (ii)−(iii):  b−a=u    ...(iv)  (i)+(iv):  2b=−u  b=−(u/2)  c=2u−b=2u+(u/2)=((5u)/2)  a=−2u−b=−2u+(u/2)=−((3u)/2)  u=a×b×c=(((−3u)/2))×(((−u)/2))×(((5u)/2))=((15)/8)u^3   (((15)/8)u^2 −1)u=0  u≠0  ⇒((15)/8)u^2 −1=0  u^2 =(8/(15))  ⇒a×b×c=u=±(√(8/(15)))=±2(√(2/(15)))
letu=a×b×ca+b=2u(i)b+c=2u(ii)a+c=u(iii)(ii)(iii):ba=u(iv)(i)+(iv):2b=ub=u2c=2ub=2u+u2=5u2a=2ub=2u+u2=3u2u=a×b×c=(3u2)×(u2)×(5u2)=158u3(158u21)u=0u0158u21=0u2=815a×b×c=u=±815=±2215

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