Question

$\frac{3\times {27}^{x+1}}{3\times 3^{3x+2}}+\frac{27\times 3^{3x}}{\frac{-1}{3}\times {27}^{x+1}}$ is equal to ____.

Answer 1

Given, $\frac{3\times {27}^{x+1}}{3\times 3^{3x+2}}+\frac{27\times 3^{3x}}{\frac{-1}{3}\times {27}^{x+1}}$

$27$ could be written as $3^3$

and $\frac{-1}{3}={-3}^{-1}$ [$\because a^{-1}=\frac{1}{a}$]

So, $\frac{3\times 3^{3(x+1)}}{3^{3x+2+1}}+\frac{3^3\times 3^{3x}}{{-3}^{-1}\times 3^{3(x+1)}}$

$=\frac{3^{3x+3+1}}{3^{3x+3}}+\frac{3^{3x+3}}{-\left(3^{3x+3-1}\right)}$ [$\because a^m\times a^n=a^{(m+n)}$]

$=\frac{3^{3x+4}}{3^{3x+3}}+\frac{3^{3x+3}}{-(3{)}^{3x+2}}$

$=3^{3x+4-(3x+3)}-(3{)}^{3x+3-(3x+2)}$ [$\because a^m÷a^n=a^{(m-n)}$]

$=3^1-3^1$

$=3-3=0$