Question

$$16x^{5}-4i^{-2}x^{3}\times(2x^{2})^{-3}; 16x^{5}-4^{-2}x^{3}\times(2x^{2})^{-3}$$

Answer

$$16*x^5-4*IM^-2*x^3*1/(8*x^6);16*x^5-x^-3/128$$

Solution


Use Negative Power Rule: \({x}^{-a}=\frac{1}{{x}^{a}}\).
\[\begin{aligned}&16{x}^{5}-4{\imath }^{-2}{x}^{3}{(2{x}^{2})}^{-3}\\&16{x}^{5}-\frac{1}{{4}^{2}}{x}^{3}{(2{x}^{2})}^{-3}\end{aligned}\]
Simplify  \({4}^{2}\)  to  \(16\).
\[\begin{aligned}&16{x}^{5}-4{\imath }^{-2}{x}^{3}{(2{x}^{2})}^{-3}\\&16{x}^{5}-\frac{1}{16}{x}^{3}{(2{x}^{2})}^{-3}\end{aligned}\]
Use Negative Power Rule: \({x}^{-a}=\frac{1}{{x}^{a}}\).
\[\begin{aligned}&16{x}^{5}-4{\imath }^{-2}{x}^{3}\times \frac{1}{{(2{x}^{2})}^{3}}\\&16{x}^{5}-\frac{1}{16}{x}^{3}\times \frac{1}{{(2{x}^{2})}^{3}}\end{aligned}\]
Use Multiplication Distributive Property: \({(xy)}^{a}={x}^{a}{y}^{a}\).
\[\begin{aligned}&16{x}^{5}-4{\imath }^{-2}{x}^{3}\times \frac{1}{{2}^{3}{({x}^{2})}^{3}}\\&16{x}^{5}-\frac{1}{16}{x}^{3}\times \frac{1}{{2}^{3}{({x}^{2})}^{3}}\end{aligned}\]
Simplify  \({2}^{3}\)  to  \(8\).
\[\begin{aligned}&16{x}^{5}-4{\imath }^{-2}{x}^{3}\times \frac{1}{8{({x}^{2})}^{3}}\\&16{x}^{5}-\frac{1}{16}{x}^{3}\times \frac{1}{8{({x}^{2})}^{3}}\end{aligned}\]
Use Power Rule: \({({x}^{a})}^{b}={x}^{ab}\).
\[\begin{aligned}&16{x}^{5}-4{\imath }^{-2}{x}^{3}\times \frac{1}{8{x}^{6}}\\&16{x}^{5}-\frac{1}{16}{x}^{3}\times \frac{1}{8{x}^{6}}\end{aligned}\]
Simplify  \(\frac{1}{16}{x}^{3}\times \frac{1}{8{x}^{6}}\)  to  \(\frac{{x}^{3}}{128{x}^{6}}\).
\[\begin{aligned}&16{x}^{5}-4{\imath }^{-2}{x}^{3}\times \frac{1}{8{x}^{6}}\\&16{x}^{5}-\frac{{x}^{3}}{128{x}^{6}}\end{aligned}\]
Use Quotient Rule: \(\frac{{x}^{a}}{{x}^{b}}={x}^{a-b}\).
\[\begin{aligned}&16{x}^{5}-4{\imath }^{-2}{x}^{3}\times \frac{1}{8{x}^{6}}\\&16{x}^{5}-\frac{{x}^{3-6}}{128}\end{aligned}\]
Simplify  \(3-6\)  to  \(-3\).
\[\begin{aligned}&16{x}^{5}-4{\imath }^{-2}{x}^{3}\times \frac{1}{8{x}^{6}}\\&16{x}^{5}-\frac{{x}^{-3}}{128}\end{aligned}\]