Question

$$\frac{(a^{3})^{n+1}\cdot(ab^{2})^{-2n}}{(ab)^{n}\cdot b^{-5n}}$$

Answer

$$a^3$$

Solution


Use Power Rule: \({({x}^{a})}^{b}={x}^{ab}\).
\[\frac{{a}^{3(n+1)}{(a{b}^{2})}^{-2n}}{{(ab)}^{n}{b}^{-5n}}\]
Use Negative Power Rule: \({x}^{-a}=\frac{1}{{x}^{a}}\).
\[\frac{{a}^{3(n+1)}\times \frac{1}{{(a{b}^{2})}^{2n}}}{{(ab)}^{n}{b}^{-5n}}\]
Use Multiplication Distributive Property: \({(xy)}^{a}={x}^{a}{y}^{a}\).
\[\frac{{a}^{3(n+1)}\times \frac{1}{{a}^{2n}{({b}^{2})}^{2n}}}{{(ab)}^{n}{b}^{-5n}}\]
Use Power Rule: \({({x}^{a})}^{b}={x}^{ab}\).
\[\frac{{a}^{3(n+1)}\times \frac{1}{{a}^{2n}{b}^{4n}}}{{(ab)}^{n}{b}^{-5n}}\]
Simplify  \({a}^{3(n+1)}\times \frac{1}{{a}^{2n}{b}^{4n}}\)  to  \(\frac{{a}^{3(n+1)}}{{a}^{2n}{b}^{4n}}\).
\[\frac{\frac{{a}^{3(n+1)}}{{a}^{2n}{b}^{4n}}}{{(ab)}^{n}{b}^{-5n}}\]
Use Quotient Rule: \(\frac{{x}^{a}}{{x}^{b}}={x}^{a-b}\).
\[\frac{{a}^{3(n+1)-2n}{b}^{-4n}}{{(ab)}^{n}{b}^{-5n}}\]
Expand.
\[\frac{{a}^{3n+3-2n}{b}^{-4n}}{{(ab)}^{n}{b}^{-5n}}\]
Collect like terms.
\[\frac{{a}^{(3n-2n)+3}{b}^{-4n}}{{(ab)}^{n}{b}^{-5n}}\]
Simplify  \((3n-2n)+3\)  to  \(n+3\).
\[\frac{{a}^{n+3}{b}^{-4n}}{{(ab)}^{n}{b}^{-5n}}\]
Use Multiplication Distributive Property: \({(xy)}^{a}={x}^{a}{y}^{a}\).
\[\frac{{a}^{n+3}{b}^{-4n}}{{a}^{n}{b}^{n}{b}^{-5n}}\]
Use Product Rule: \({x}^{a}{x}^{b}={x}^{a+b}\).
\[\frac{{a}^{n+3}{b}^{-4n}}{{a}^{n}{b}^{n-5n}}\]
Simplify  \(n-5n\)  to  \(-4n\).
\[\frac{{a}^{n+3}{b}^{-4n}}{{a}^{n}{b}^{-4n}}\]
Cancel \({b}^{-4n}\).
\[\frac{{a}^{n+3}}{{a}^{n}}\]
Use Quotient Rule: \(\frac{{x}^{a}}{{x}^{b}}={x}^{a-b}\).
\[{a}^{n+3-n}\]
Collect like terms.
\[{a}^{(n-n)+3}\]
Simplify  \((n-n)+3\)  to  \(3\).
\[{a}^{3}\]