Question

$a^3+3a^{2}b+3a{b}^2+b^3-8$

• A. $(a-b-2)(a^2+b^2-2ab+4a+2a+2b)$
• B. $(a+b{)}^3$
• C. $(a-b{)}^3$
• D. $(a+b-2)(a^2+b^2+2ab+4a+2a+2b)$

Answer 1

The correct option is D $(a+b-2)(a^2+b^2+2ab+4a+2a+2b)$

We solve the given expression $a^3+3a^{2}b+3a{b}^2+b^3-8$ as shown below:

$a^3+3a^{2}b+3a{b}^2+b^3-8=(a+b{)}^3-(2{)}^3(\because (x+y{)}^3=x^3+y^3+3x^{2}y+3x{y}^2)=(a+b-2\left)\right[(a+b{)}^2+2^2+2(a+b\left)\right](\because x^3-y^3=(x-y\left)\right(x^2+y^2+xy\left)\right)=(a+b-2)(a^2+b^2+2ab+4+2a+2b)(\because (x+y{)}^2=x^2+y^2+xy)$

Hence, $a^3+3a^{2}b+3a{b}^2+b^3-8=(a+b-2)(a^2+b^2+2ab+4+2a+2b)$