A^2=b+c,b^2=c+a,c^2=a+b, then the value of 1/a+b + 1/b+1 + 1/c+1 equal to?

1 Answer
May 17, 2018

# 3, or, 1#.

Explanation:

We will assume that, #a,b,c in RR#.

We have, #a^2=b+c, and, b^2=c+a#.

#:. a^2-b^2=(b+c)-(c+a)=b-a#.

#:. (a-b)(a+b)+(a-b)=0#.

#:. (a-b)(a+b+1)=0#.

#:. a=b, or, a+b+1=0#.

#"If "a+b+1=0," then, "c^2=a+b=-1," which is impossible in "RR#.

#:. a=b#, and, we can similarly get, #b=c#.

#:. a=b=c#.

Then, #a^2=b+c rArr a^2=a+a=2a rArr a^2-2a=0#.

#:. a=0, or, a=2#.

Thus, #a=b=c=0, or, a=b=c=2#.

Case 1 : #a=b=c=0#.

In this case, #1/(a+1)+1/(b+1)+1/(c+1)=1/1+1/1+1/1=3#.

Case 2 : #a=b=c=2#.

In this case, the Reqd. Value #=3*1/3=1#.