Q. 161 - Intermediate Algebra

Question

In the following exercises,simplify

$$\begin{gathered} \left(a\right) \frac{c^{{\displaystyle \frac{5}{3}}}.c^{-{\displaystyle \frac{1}{3}}}}{c^{-{\displaystyle \frac{2}{3}}}} \\ \left(b\right) {\left(\frac{8 x^{{\displaystyle \frac{5}{3}}}.y^{-{\displaystyle \frac{1}{2}}}}{27 x^{-{\displaystyle \frac{4}{3}}}.y^{{\displaystyle \frac{5}{2}}}}\right)}^{\frac{1}{3}} \end{gathered}$$

Step-by-Step Solution

Answer

 $$\begin{gathered} \left(a\right) c^2 \\ \left(b\right) \frac{2x}{3y} \end{gathered}$$

Part (a) Step 1.Given information

Consider the given expression,$\frac{c^{{\displaystyle \frac{5}{3}}}.c^{-{\displaystyle \frac{1}{3}}}}{c^{-{\displaystyle \frac{2}{3}}}}$

To simplify the expression,by used product and quotient property that is,

$a^m.a^n=a^{m+n} and \left(\frac{a^m}{a^n}\right)=a^{m-n}$

Part (a) Step 2.Simplify the expression

$$\begin{gathered} \frac{c^{{\displaystyle \frac{5}{3}}}.c^{-{\displaystyle \frac{1}{3}}}}{c^{-{\displaystyle \frac{2}{3}}}}=\frac{c^{{\displaystyle \frac{5}{3}}+\left(-{\displaystyle \frac{1}{3}}\right)}}{c^{-{\displaystyle \frac{2}{3}}}} [Since ; a^m.a^n=a^{m+n}] \\ =\frac{c^{{\displaystyle \frac{4}{3}}}}{c^{-{\displaystyle \frac{2}{3}}}} [Add the exponents] \\ = c^{\frac{4}{3}-\left(-\frac{2}{3}\right) [Since ;\frac{a^m}{a^n}=a^{m-n}]} \\ =c^{\frac{4}{3}+\frac{2}{3}} [Simplify the exponents] \\ =c^2 \left[Simplify\right] \\ Therefore,the answer is c^2 \end{gathered}$$

Part (b) Step 1.Given information

Consider the given expression,${\left(\frac{8 x^{{\displaystyle \frac{5}{3}}}.y^{-{\displaystyle \frac{1}{2}}}}{27 x^{-{\displaystyle \frac{4}{3}}}.y^{{\displaystyle \frac{5}{2}}}}\right)}^{\frac{1}{3}}$

To simplify the expression,by used product and quotient property that is,$a^m.a^n=a^{m+n} and \left(\frac{a^m}{a^n}\right)=a^{m-n}$

Part (b) Step 2.Simplify the expression

$$\begin{gathered} {\left(\frac{8 x^{{\displaystyle \frac{5}{3}}}.y^{-{\displaystyle \frac{1}{2}}}}{27 x^{-{\displaystyle \frac{4}{3}}}.y^{{\displaystyle \frac{5}{2}}}}\right)}^{\frac{1}{3}}={\left(\frac{8}{27}\right)}^{\frac{1}{3}}.{\left(x^{\frac{5}{3}-\left(-\frac{4}{3}\right)}\right)}^{\frac{1}{3}}.{\left(y^{-\frac{1}{2}-\frac{5}{2}}\right)}^{\frac{1}{3}} [Since ; \frac{a^m}{a^n}=a^{m-n}] \\ ={\left[{\left(\frac{2}{3}\right)}^3\right]}^{\frac{1}{3}}.{\left(x^{\frac{5}{3}-\left(-\frac{4}{3}\right)}\right)}^{\frac{1}{3}}.{\left(y^{-\frac{1}{2}-\frac{5}{2}}\right)}^{\frac{1}{3}} [Since ; \left(\frac{a^m}{a^n}\right)=a^{m-n}] \\ ={\left(\frac{2}{3}\right)}^{3\frac{1}{3}}.{\left(x^3\right)}^{\frac{1}{3}}.{\left(y^{-3}\right)}^{\frac{1}{3}} [Substract the exponent] \\ ={\left(\frac{2}{3}\right)}^{\frac{3}{3}}.x^{\frac{3}{3}}.y^{\frac{-3}{3}} [Since ;{\left(a^m\right)}^n=a^{m.n}] \\ =\frac{2}{3}.x.y^{-1 [Divide the exponent]} \\ = \frac{2x}{3y} [Since ; a^{-n}=\frac{1}{a^n}] \\ Therefore,the answer is \frac{2x}{3y} \end{gathered}$$