Question

$$\frac{(3xy)^{2}\times\sqrt{x^{4}y^{6}}}{(2x^{4}y^{3})^{-1}}$$

Answer

$$18*x^8*y^8$$

Solution


Use this rule: \(\sqrt{ab}=\sqrt{a}\sqrt{b}\).
\[\frac{{(3xy)}^{2}\sqrt{{x}^{4}}\sqrt{{y}^{6}}}{{(2{x}^{4}{y}^{3})}^{-1}}\]
Simplify  \(\sqrt{{x}^{4}}\)  to  \({x}^{2}\).
\[\frac{{(3xy)}^{2}{x}^{2}\sqrt{{y}^{6}}}{{(2{x}^{4}{y}^{3})}^{-1}}\]
Simplify  \(\sqrt{{y}^{6}}\)  to  \({y}^{3}\).
\[\frac{{(3xy)}^{2}{x}^{2}{y}^{3}}{{(2{x}^{4}{y}^{3})}^{-1}}\]
Use Multiplication Distributive Property: \({(xy)}^{a}={x}^{a}{y}^{a}\).
\[\frac{{3}^{2}{x}^{2}{y}^{2}{x}^{2}{y}^{3}}{{(2{x}^{4}{y}^{3})}^{-1}}\]
Simplify  \({3}^{2}\)  to  \(9\).
\[\frac{9{x}^{2}{y}^{2}{x}^{2}{y}^{3}}{{(2{x}^{4}{y}^{3})}^{-1}}\]
Regroup terms.
\[\frac{9{x}^{2}{x}^{2}{y}^{2}{y}^{3}}{{(2{x}^{4}{y}^{3})}^{-1}}\]
Use Product Rule: \({x}^{a}{x}^{b}={x}^{a+b}\).
\[\frac{9{x}^{2+2}{y}^{2+3}}{{(2{x}^{4}{y}^{3})}^{-1}}\]
Simplify  \(2+2\)  to  \(4\).
\[\frac{9{x}^{4}{y}^{2+3}}{{(2{x}^{4}{y}^{3})}^{-1}}\]
Simplify  \(2+3\)  to  \(5\).
\[\frac{9{x}^{4}{y}^{5}}{{(2{x}^{4}{y}^{3})}^{-1}}\]
Use Negative Power Rule: \({x}^{-a}=\frac{1}{{x}^{a}}\).
\[\frac{9{x}^{4}{y}^{5}}{\frac{1}{2{x}^{4}{y}^{3}}}\]
Invert and multiply.
\[9{x}^{4}{y}^{5}\times 2{x}^{4}{y}^{3}\]
Take out the constants.
\[(9\times 2){x}^{4}{x}^{4}{y}^{5}{y}^{3}\]
Simplify  \(9\times 2\)  to  \(18\).
\[18{x}^{4}{x}^{4}{y}^{5}{y}^{3}\]
Use Product Rule: \({x}^{a}{x}^{b}={x}^{a+b}\).
\[18{x}^{4+4}{y}^{5+3}\]
Simplify  \(4+4\)  to  \(8\).
\[18{x}^{8}{y}^{5+3}\]
Simplify  \(5+3\)  to  \(8\).
\[18{x}^{8}{y}^{8}\]