Question : If $a+b+c$ = 6 and $ab+bc+ca$ = 11, then the value of $bc(b+c)+ca(c+a)+ab(a+b)+3abc$ is:
Option 1: 33
Option 2: 66
Option 3: 55
Option 4: 23
Answer (1)
22nd Jan, 2024
Correct Answer: 66
Solution :
Given: $a+b+c$ = 6 and $ab+bc+ca$ = 11
To find:
$bc(b+c)+ca(c+a)+ab(a+b)+3abc$
= $b^2c+bc^2+c^2a+ca^2+a^2b+ab^2+abc+abc+abc$
= $bc(a+b+c)+ac(a+b+c)+ab(a+b+c)$
= $(a+b+c)(ab+bc+ac)$
Putting the values, we get
⇒ 6$×$11 = 66
Hence, the correct answer is 66.
Comments (0)
Related Questions
Know More about
Staff Selection Commission Combined Grad ...
Result | Eligibility | Application | Selection Process | Preparation Tips | Admit Card | Answer Key
Get Updates Brochure
Thank You!
Your Staff Selection Commission Combined Graduate Level Exam brochure has been successfully mailed to your registered email id “”.
Upcoming Exams
Application Date:
9 Dec, 2024
- 30 Dec, 2024
Application Date:
2 Dec, 2024
- 31 Dec, 2024
Exam Date:
3 Jan, 2025
- 16 Jan, 2025
Admit Card Date:
23 Dec, 2024
- 5 Jan, 2025
Application Date:
15 Dec, 2024
- 5 Jan, 2025
