Question

The value of $(a+b)(a+c)(b+c)$, where $a=1$, $b=2$ and $c=3$ will be .

• A. 46
• B. 60
• C. 50

Answer 1

The correct option is B 60

Given: $(a+b)(a+c)(b+c)$
To find : The value of $(a+b)(a+c)(b+c)$ when $a=1$, $b=2$ and $c=3$
By applying distributivity property,
$(a+b)(a+c)(b+c)=(a^2+ac+ba+bc)(b+c)=a^{2}b+a^{2}c+bac+a{c}^2+a{b}^2+abc+b^{2}c+b{c}^2$

On substituting the values $a=1$, $b=2$ and $c=3$, we get,
$(1^2\times 2)+(1^2\times 3)+(2\times 1\times 3)+(1\times 3^2)+(1\times 2^2)+(1\times 2\times 3)+(2^2\times 3)(2\times 3^2)=2+3+6+9+4+6+12+18=60$