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
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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.
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