If 1 a 1 b 1 c 1 a b c where a b c is not equal to 0

let a= 1, b= -1 and c= 2
verifing a+b+c is not equals to zero. and 1/a + 1/b + 1/c = 1/(a+b+c)
So, (a+b)(b+c)(c+a)= 0
10 years agoHelpfull: Yes(23) No(1)
Ans is o i.e, (a)
As 1/a+1/b+1/c=1/(a+b+c),
=>(a+b+c)(ab+bc+ac)=abc
=>a^2(b+c)+b^2(a+c)+c^2(a+b)-2abc=0.
Similarly on solving (a+b)(b+c)(c+a)=a^2(b+c)+b^2(a+c)+c^2(a+b)=0
10 years agoHelpfull: Yes(11) No(0)
Answer is 0.
1/a + 1/b + 1/c =1/ (a+b+c)
=> 1/a + 1/b =1/ (a+b+c) - 1/c
=> (b+a)/ab = (c - (a+b+c)) / c(a+b+c)

Cross multiplying (Since, a,b,c !=0 and a+b+c != 0)

=> c(a+b+c)(b+a) = - ab(a+b)
=> (a+b)( ca +cb +cc + ab ) = 0
=> (a+b)( ca + cc + cb + ab) = 0
=> (a+b)(b+c)(c+a)=0
9 years agoHelpfull: Yes(4) No(0)
it's simple ANS is :0
now assume a=-1,b=1,c=1
then If 1/a + 1/b + 1/c = 1/(a+b+c) is:1/-1 +1/1 +1/1=1/(-1+1+1)
:1=1
condition 1: a+b+c!=0 ,-1+1+1=1!=0 satisfied
(a+b)(b+c)(c+a)=(-1+1)(1+1)(1-1)=0
so ANS is 0

so
10 years agoHelpfull: Yes(2) No(0)
It must be 0 na???
Plz let me know if u have different options ...
(c)
10 years agoHelpfull: Yes(1) No(0)
ans is 0
10 years agoHelpfull: Yes(1) No(2)
(a) 1/a+1/b+1/c=1/(a+b+c) =>(bc+ac+ab)/abc=1/(a+b+c)
=>0
10 years agoHelpfull: Yes(1) No(0)
Ans :0
by simplifying The Equation we can Get the Result
10 years agoHelpfull: Yes(0) No(0)
c) 1
since (a+b)(b+c)(c+a)=(1/a+ 1/b + 1/c)(a+b+c)
u can check it.
10 years agoHelpfull: Yes(0) No(1)

If 1/a + 1/b + 1/c = 1/(a+b+c)where a+b+c is not equal to 0, what is the value of (a+b)(b+c)(c+a)...
(a) 0
(b) >0
(c) 1