Question

Simplify the following using laws of exponent
$\frac{(7xyz{)}^0}{\left[12x^2\right(y^8{)}^{-3}(z^3{)}^2{]}^{-1}}$

Answer 1

We have,

$=\frac{(7xyz{)}^0}{\left[12x^2\right(y^8{)}^{-3}(z^3{)}^2{]}^{-1}}$

$=\frac{1}{\left[12x^2\right(y^8{)}^{-3}(z^3{)}^2{]}^{-1}}$

$=12x^2\left(y^8{)}^{-3}\right(z^3{)}^2$

$=12x^2{y}^{-24}{z}^6$

$=\frac{12x^2{z}^6}{y^{24}}$

Hence, this is the answer.