Question

1/a+b+c=1/a+1/b+1/c​

Answer 1

Answer:

Step-by-step explanation:

1a+1b+1c=1a+b+c

⟹1a+1b=1a+b+c−1c

⟹a+bab=c−(a+b+c)c(a+b+c)

⟹a+bab=−(a+b)c(a+b+c)

Conclusions:

Case 1:

(a+b)=0

⟹b=−a

[You may also get (b+c)=0 or (c+a)=0 depending on how we group the terms.]

Case 2:

(a+b)≠0

We can continue we our earlier calculation.

a+bab=−(a+b)c(a+b+c)

⟹1ab=−1c(a+b+c)

⟹1ab+1c(a+b+c)=0

⟹c(a+b+c)+ab=0

⟹c2=−(ab+bc+ca)  

It can shown that

a2=−(ab+bc+ca)

b2=−(ab+bc+ca)

c2=−(ab+bc+ca)

This implies a2=b2=c2=k2

(where 'k' is a positive constant.

a=±k

b=±k

c=±k

Answer 2

I hope it is helpful for u