Question

please koi batao please​

Answer 1

take the lcm and put the value ab, bc ca from above eqation in numerator and denominator you will get answer 0

Answer 2

Answer:

(4) 0

Step-by-step explanation:

Given :

(a,b,c) & (a+b+c) ≠ 0

And, ab+bc+ca = 0

consider the following :

bc + ca = -ab (1)

ab + ca = -bc (2)

ab + bc = -ca (3)

Now the given question reduces to :

(1/(a²+(ab+ca))) + (1/(b²+(ab+bc))) + (1/(c²+(bc+ca)))

=> Taking a,b,c common in respective denominators

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

Taking 1/(a+b+c) common

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

=> Simplifying the second Term

=> (bc + ca + ab)/(abc)

But, According to the given data the numerator is 0, => The whole fraction is 0 => With the left term being multiplied to 0 , The answer is 0

Therefore the answer is (4)0

If you have doubts still, Pls comment.

Have a great day,

Thanking you,

Bunti 360 !