Question

$$(\frac{1}{x^{0}x^{-a}})^{b-c}\times(\frac{4x^{-b}}{16\div4})^{a-c}\times(\frac{1}{x^{c}})^{b-a}$$

Answer

1

Solution


Use Rule of Zero: \({x}^{0}=1\).
\[{(\frac{1}{1\times {x}^{-a}})}^{b-c}{(\frac{4{x}^{-b}}{\frac{16}{4}})}^{a-c}{(\frac{1}{{x}^{c}})}^{b-a}\]
Simplify  \(1\times {x}^{-a}\)  to  \({x}^{-a}\).
\[{(\frac{1}{{x}^{-a}})}^{b-c}{(\frac{4{x}^{-b}}{\frac{16}{4}})}^{a-c}{(\frac{1}{{x}^{c}})}^{b-a}\]
Simplify  \(\frac{16}{4}\)  to  \(4\).
\[{(\frac{1}{{x}^{-a}})}^{b-c}{(\frac{4{x}^{-b}}{4})}^{a-c}{(\frac{1}{{x}^{c}})}^{b-a}\]
Cancel \(4\).
\[{(\frac{1}{{x}^{-a}})}^{b-c}{({x}^{-b})}^{a-c}{(\frac{1}{{x}^{c}})}^{b-a}\]
Use Division Distributive Property: \({(\frac{x}{y})}^{a}=\frac{{x}^{a}}{{y}^{a}}\).
\[\frac{1}{{({x}^{-a})}^{b-c}}{({x}^{-b})}^{a-c}{(\frac{1}{{x}^{c}})}^{b-a}\]
Use Power Rule: \({({x}^{a})}^{b}={x}^{ab}\).
\[\frac{1}{{x}^{-a(b-c)}}{({x}^{-b})}^{a-c}{(\frac{1}{{x}^{c}})}^{b-a}\]
Use Power Rule: \({({x}^{a})}^{b}={x}^{ab}\).
\[\frac{1}{{x}^{-a(b-c)}}{x}^{-b(a-c)}{(\frac{1}{{x}^{c}})}^{b-a}\]
Use Division Distributive Property: \({(\frac{x}{y})}^{a}=\frac{{x}^{a}}{{y}^{a}}\).
\[\frac{1}{{x}^{-a(b-c)}}{x}^{-b(a-c)}\times \frac{1}{{({x}^{c})}^{b-a}}\]
Use Power Rule: \({({x}^{a})}^{b}={x}^{ab}\).
\[\frac{1}{{x}^{-a(b-c)}}{x}^{-b(a-c)}\times \frac{1}{{x}^{c(b-a)}}\]
Use this rule: \(\frac{a}{b} \times \frac{c}{d}=\frac{ac}{bd}\).
\[\frac{1\times {x}^{-b(a-c)}\times 1}{{x}^{-a(b-c)}{x}^{c(b-a)}}\]
Simplify  \(1\times {x}^{-b(a-c)}\times 1\)  to  \({x}^{-b(a-c)}\).
\[\frac{{x}^{-b(a-c)}}{{x}^{-a(b-c)}{x}^{c(b-a)}}\]
Use Product Rule: \({x}^{a}{x}^{b}={x}^{a+b}\).
\[\frac{{x}^{-b(a-c)}}{{x}^{-a(b-c)+c(b-a)}}\]
Expand.
\[\frac{{x}^{-b(a-c)}}{{x}^{-ab+ac+cb-ca}}\]
Collect like terms.
\[\frac{{x}^{-b(a-c)}}{{x}^{-ab+(ac-ac)+cb}}\]
Simplify  \(-ab+(ac-ac)+cb\)  to  \(-ab+cb\).
\[\frac{{x}^{-b(a-c)}}{{x}^{-ab+cb}}\]
Use Quotient Rule: \(\frac{{x}^{a}}{{x}^{b}}={x}^{a-b}\).
\[{x}^{-b(a-c)-(-ab+cb)}\]
Remove parentheses.
\[{x}^{-b(a-c)+ab-cb}\]
Expand.
\[{x}^{-ba+bc+ab-cb}\]
Collect like terms.
\[{x}^{-ba+(bc-bc)+ab}\]
Simplify  \(-ba+(bc-bc)+ab\)  to  \(-ba+ab\).
\[{x}^{-ba+ab}\]
Simplify.
\[{x}^{-ab+ab}\]
Simplify  \(-ab+ab\)  to  \(0\).
\[{x}^{0}\]
Use Rule of Zero: \({x}^{0}=1\).
\[1\]