Question

If 1/a+1/b+1/c = 1/(a+b+c) where a+b+c not equals '0', abc not equals '0', What is the value of (a+b) (b+c) (a+c) ?

A) 3 B) 2 C) 1 D) 0

Answer 1

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

=> 1/a + 1/b = 1/(a+b+c) - 1/c

=> ( a + b ) / ab = {c - ( a+b+c ) } / c (a+b+c)

=> ( a+b ) / ab = { c - a - b - c } / c (a+b+c)

=> ( a+b ) / ab = - ( a+b ) / c (a+b+c)

=> ( a+b ) [ 1/ab + 1 / (a+b+c) c ] = 0

=> a+b= 0

therefore ,

(a+b)(b+c)(c+a) = 0 ....... AnSwEr

Answer 2

the correct answer is option (d) 0

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