Question

$$\frac{(x^{a+b})^{a-b}\times(x^{b+c})^{b-c}}{(x^{a+c})^{a-c}}$$

Answer

1

Solution


Use Power Rule: \({({x}^{a})}^{b}={x}^{ab}\).
\[\frac{{x}^{(a+b)(a-b)}{({x}^{b+c})}^{b-c}}{{({x}^{a+c})}^{a-c}}\]
Use Power Rule: \({({x}^{a})}^{b}={x}^{ab}\).
\[\frac{{x}^{(a+b)(a-b)}{x}^{(b+c)(b-c)}}{{({x}^{a+c})}^{a-c}}\]
Use Product Rule: \({x}^{a}{x}^{b}={x}^{a+b}\).
\[\frac{{x}^{(a+b)(a-b)+(b+c)(b-c)}}{{({x}^{a+c})}^{a-c}}\]
Expand.
\[\frac{{x}^{{a}^{2}-{b}^{2}+{b}^{2}-{c}^{2}}}{{({x}^{a+c})}^{a-c}}\]
Collect like terms.
\[\frac{{x}^{{a}^{2}+(-{b}^{2}+{b}^{2})-{c}^{2}}}{{({x}^{a+c})}^{a-c}}\]
Simplify  \({a}^{2}+(-{b}^{2}+{b}^{2})-{c}^{2}\)  to  \({a}^{2}-{c}^{2}\).
\[\frac{{x}^{{a}^{2}-{c}^{2}}}{{({x}^{a+c})}^{a-c}}\]
Use Power Rule: \({({x}^{a})}^{b}={x}^{ab}\).
\[\frac{{x}^{{a}^{2}-{c}^{2}}}{{x}^{(a+c)(a-c)}}\]
Use Quotient Rule: \(\frac{{x}^{a}}{{x}^{b}}={x}^{a-b}\).
\[{x}^{{a}^{2}-{c}^{2}-(a+c)(a-c)}\]
Expand.
\[{x}^{{a}^{2}-{c}^{2}-{a}^{2}+{c}^{2}}\]
Collect like terms.
\[{x}^{({a}^{2}-{a}^{2})+(-{c}^{2}+{c}^{2})}\]
Simplify  \(({a}^{2}-{a}^{2})+(-{c}^{2}+{c}^{2})\)  to  \(0\).
\[{x}^{0}\]
Use Rule of Zero: \({x}^{0}=1\).
\[1\]