Question

Solve this:
Q. Simplify each expression. Use positive exponents.
1. $m^3{n}^{-6}{p}^0$
2. $\frac{a^4{b}^{-3}}{a{b}^{-2}}$
3.${\left(x^{-2}{y}^{-4}{x}^3\right)}^{-2}$

Answer 1

Dear Student,
Here is the solution of your asked query:
$$\begin{gathered} 1){\mathrm{m}}^3{\mathrm{n}}^{-6}{\mathrm{p}}^0 \\ ={\mathrm{m}}^3\times \frac{1}{{\mathrm{n}}^6}\times 1\{\mathrm{since}{\mathrm{a}}^{-\mathrm{m}}=\frac{1}{{\mathrm{a}}^{\mathrm{m}}}\mathrm{and}{\mathrm{a}}^0=1\} \\ =\frac{{\mathrm{m}}^3}{{\mathrm{n}}^6} \\ 2)\frac{{\mathrm{a}}^4{\mathrm{b}}^{-3}}{{\mathrm{ab}}^{-2}} \\ =\frac{{\mathrm{a}}^4}{\mathrm{a}}\times \frac{{\mathrm{b}}^{-3}}{{\mathrm{b}}^{-2}} \\ ={\mathrm{a}}^{4-1}\times {\mathrm{b}}^{-3-\left(-2\right)}\{\mathrm{Since}\frac{{\mathrm{a}}^{\mathrm{m}}}{{\mathrm{a}}^{\mathrm{n}}}={\mathrm{a}}^{\mathrm{m}-\mathrm{n}}\} \\ ={\mathrm{a}}^3\times {\mathrm{b}}^{-3+2} \\ ={\mathrm{a}}^3\times {\mathrm{b}}^{-1} \\ =\frac{{\mathrm{a}}^3}{\mathrm{b}}\{\mathrm{since}{\mathrm{x}}^{-1}=\frac{1}{\mathrm{x}}\} \\ 3){\left({\mathrm{x}}^{-2}\times {\mathrm{y}}^{-4}\times {\mathrm{x}}^3\right)}^{-2} \\ ={\left({\mathrm{x}}^{-2}\right)}^{-2}\times {\left({\mathrm{y}}^{-4}\right)}^{-2}\times {\left({\mathrm{x}}^3\right)}^{-2} \\ ={\mathrm{x}}^{-2\times \left(-2\right)}\times {\mathrm{y}}^{\left(-4\right)\times \left(-2\right)}\times {\mathrm{x}}^{3\times \left(-2\right)} \\ ={\mathrm{x}}^4\times {\mathrm{y}}^8\times {\mathrm{x}}^{-6} \\ =x^{4-6}\times y^8 \\ =x^{-2}\times y^8 \\ =\frac{y^8}{x^2} \end{gathered}$$
Regards