if 1 a 1 b 1 c a b c a b c abc is not equal to zero then

d)can not be determined
bcoz if we put the value of a,b,c 1,1,-1 ans is zero
if we put the value of a,b,c 1,1,1 ans is zero is 8
so ans cab be 0 0r 8 so can not be determined
10 years agoHelpfull: Yes(11) No(1)

d)can not be determined

Because when we put 1,1,1 for a,b and c the ans is > 0,
but when we put 1,1,-1 for a,b and c the ans is = 0,
when we put -1,-1,-1 for a,b and c the ans is < 0. Therefore the ans is d.
10 years agoHelpfull: Yes(7) No(1)
It can only be possible when all are equal to 1 because no other value can satisfy the given condition. Therefore, the answer is 8.
10 years agoHelpfull: Yes(4) No(8)
C) Greater then zero

Because a=1,b=1,c=1 satisfies the equation 1/a+1/b+... and we substitute the values in (a+b)(b+c)(c+a) value is 8.
10 years agoHelpfull: Yes(3) No(3)
(a+b)(b+c)(c+a)=a2b+a2c+b2a+b2c+c2a+c2b+2abc.........(1)
now, solve
wll get..
a2b+a2c+b2a+b2c+c2a+c2b+2abc=0...(2)
from (1) & (2)
(a+b)(b+c)(c+a)=0
10 years agoHelpfull: Yes(2) No(19)
ans is 0,just solve 1/a+1/b+1/c=a+b+c
10 years agoHelpfull: Yes(2) No(2)
(1,1,1) satisfies and gives >0 but
(1,1,-1) also satisfies and gives 0. so please correct me here..
10 years agoHelpfull: Yes(1) No(0)
greater than 0
10 years agoHelpfull: Yes(0) No(1)
solving eq
Σab=abc*Σa
also
(a+b)(b+c)(c+a)=Σab*Σa-abc=>abc*(Σa^2-1)

but we will get both -ve and +ve for different values of a,b,c=(1,1,1),(-1,-1,-1)
so D
10 years agoHelpfull: Yes(0) No(0)
Just hit n trial and make sure a b c are distinct values
so a= 2 b= -2 c= 1
ans is 0
9 years agoHelpfull: Yes(0) No(0)
post correct questions only.....
9 years agoHelpfull: Yes(0) No(0)
ques is wrong correct ques is in down
9 years agoHelpfull: Yes(0) No(0)

if 1/a+1/b+1/c=a+b+c, (a+b+c),(abc)is not equal to zero,then what is the value of (a+b)(b+c)(c+a)?

a)equals to zero
b)less than zero
c)greater than zero
d)can not be determined