Question #96981

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Answer 1 · 2018-02-03T18:32:08

# 1#.

Given that,

#abc=1," we have, "1/c=ab, and, c=1/(ab)......(star)#.

#:. sum1/(1+a+1/b)#,

#=1/(1+a+1/b)+1/(1+b+1/c)+1/(1+c+1/a)#,

#=1/(1+a+1/b)+1/(1+b+ab)+1/(1+1/(ab)+1/a)......[because, (star)]#,

#=b/{b(1+a+1/b)}+1/(1+b+ab)+(ab)/{ab(1+1/(ab)+1/a)}#,

#=b/(b+ab+1)+1/(1+b+ab)+(ab)/(ab+1+b)#,

#=(b+1+ab)/(b+1+ab)#,

#=1#.