Question

Expand the algebraic expression $(3a^2+2b^3)(3a^2-2b^3).$

• A. $9a^4-4b^9$
• B. $9a^4-4b^6$
• C. $3a^4-2b^6$
• D. $3a^4-2b^9$

Answer 1

The correct option is B $9a^4-4b^6$

$\underset{–}{\text{Need to simplify:}}(3a^2+2b^3)(3a^2-2b^3)$

$(3a^2+2b^3)\times (3a^2-2b^3)$

Applying distributive property of multiplication: $(a+b)\times c=a\times c+b\times c$

$=3a^2(3a^2-2b^3)+2b^3(3a^2-2b^3)$

$=3a^2\cdot 3a^2-3a^2\cdot 2b^3+2b^3\cdot 3a^2-2b^3\cdot 2b^3$

$=3\cdot 3\cdot a^2\cdot a^2-3\cdot 2\cdot a^2\cdot b^3+2\cdot 3\cdot b^3\cdot a^2-2\cdot 2\cdot b^3\cdot b^3$

$=9a^{2+2}-6a^2{b}^3+6a^2{b}^3-4b^{3+3}$
$[\because p^m\cdot p^n=p^{m+n}]$

$=9a^4-4b^6$
$[\because -6a^2{b}^3\text{and}+6a^2{b}^3$ cancel out each other$]$

$\therefore \underset{–}{(3a^2+2b^3)(3a^2-2b^3)=9a^4-4b^6}$

Hence, option (b.) is the correct one.

$\underset{–}{\text{Note:}}$
We get:
$(3a^2+2b^3)(3a^2-2b^3)$
$=9a^4-4b^6$
$=(3a^2{)}^2-(2b^3{)}^2$
$\Rightarrow (3a^2+2b^3)(3a^2-2b^3)=(3a^2{)}^2-(2b^3{)}^2$
Hence, we can generalize: $(p+q)(p-q)=p^2-q^2$