Question

Which of the following option is equal to $3x^2{y}^3(4x^3+2x{y}^2+x^{-2}{y}^{-3})$ for $x,y\ne 0?$

• A. $12x^5{y}^3+6x^5{y}^3+3$
• B. $12x^5{y}^3+6x^3{y}^5$
• C. $12x^5{y}^3+6x^3{y}^5+3$
  
• D. $12x^3{y}^5+6x^3{y}^5+3$

Answer 1

The correct option is C $12x^5{y}^3+6x^3{y}^5+3$


$\underset{–}{\text{Need to simplify:}}3x^2{y}^3(4x^3+2x{y}^2+x^{-2}{y}^{-3})\text{for}x,y\ne 0$

$3x^2{y}^3(4x^3+2x{y}^2+x^{-2}{y}^{-3})$

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

$=3x^2{y}^3\cdot 4x^3+3x^2{y}^3\cdot 2x{y}^2+3x^2{y}^3\cdot x^{-2}{y}^{-3}$

Applying multiplication law of exponent: $a^m\cdot a^n=a^{m+n}$

$=3\cdot 4\cdot x^{2+3}\cdot y^3+3\cdot 2\cdot x^{2+1}\cdot y^{3+2}+3\cdot x^{2+(-2)}\cdot y^{3+(-3)}$

$=12x^5{y}^3+6x^3{y}^5+3x^0{y}^0$

$=12x^5{y}^3+6x^3{y}^5+3$
$[\because a^0=1$ for all $a\ne 0\Rightarrow x^0=y^0=1]$

$\therefore \underset{–}{3x^2{y}^3(4x^3+2x{y}^2+x^{-2}{y}^{-3})=12x^5{y}^3+6x^3{y}^5+3}$

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