5 Views

Question : If $a+b+c = 0$, then the value of $\small \frac{1}{(a+b)(b+c)}+\frac{1}{(b+c)(c+a)}+\frac{1}{(c+a)(a+b)}$ is:

Option 1: 0

Option 2: 1

Option 3: 3

Option 4: 2


Team Careers360 11th Jan, 2024
Answer (1)
Team Careers360 13th Jan, 2024

Correct Answer: 0


Solution : Given: $a+b+c = 0$
Solution:
$\frac{1}{(a+b)(b+c)}+\frac{1}{(b+c)(c+a)}+\frac{1}{(c+a)(a+b)}$
Taking the LCM, we get
= $\frac{(c+a)+(a+b)+(b+c)}{(a+b)(b+c)(c+a)}$
= $\frac{2(a+b+c)}{(a+b)(b+c)(c+a)}$
Substituting the value of $(a+b+c)$ in the equation:
= $\frac{2\times 0}{(a+b)(b+c)(c+a)}$
= 0
Hence, the correct answer is 0.

SSC CGL Complete Guide

Candidates can download this ebook to know all about SSC CGL.

Download EBook

Know More About

Related Questions

TOEFL ® Registrations 2024
Apply
Accepted by more than 11,000 universities in over 150 countries worldwide
Manipal Online M.Com Admissions
Apply
Apply for Online M.Com from Manipal University
GRE ® Registrations 2024
Apply
Apply for GRE® Test now & save 10% with ApplyShop Gift Card | World's most used Admission Test for Graduate & Professional Schools
View All Application Forms

Download the Careers360 App on your Android phone

Regular exam updates, QnA, Predictors, College Applications & E-books now on your Mobile

150M+ Students
30,000+ Colleges
500+ Exams
1500+ E-books