Answer
$\dfrac{a^{11/8}}{b^{1/6}}$
Work Step by Step
$\bf{\text{Solution Outline:}}$
Use the laws of exponents to simplify the given expression, $
(a^{3/4}b^{2/3})(a^{5/8}b^{-5/6})
.$
$\bf{\text{Solution Details:}}$
Using the Product Rule of the laws of exponents which is given by $x^m\cdot x^n=x^{m+n},$ the expression above is equivalent to
\begin{array}{l}\require{cancel}
a^{\frac{3}{4}+\frac{5}{8}}b^{\frac{2}{3}+\left(-\frac{5}{6} \right)}
\\\\=
a^{\frac{3}{4}+\frac{5}{8}}b^{\frac{2}{3}-\frac{5}{6}}
.\end{array}
To add $\dfrac{3}{4}$ and $\dfrac{5}{8},$ change the fractions to similar fractions (same denominator) by using the $LCD$. The $LCD$ of the denominators $4$ and $8$ is $8$ since it is the lowest number that can be exactly divided by the denominators. Multiplying the fractions by an expression equal to $1$ that will make the denominator equal to the $LCD$ results to
\begin{array}{l}\require{cancel}
\dfrac{3}{4}\cdot\dfrac{2}{2}+\dfrac{5}{8}
\\\\=
\dfrac{6}{8}+\dfrac{5}{8}
\\\\=
\dfrac{11}{8}
.\end{array}
To simplify the expression $
\dfrac{2}{3}-\dfrac{5}{6}
,$ change the fractions to similar fractions (same denominator) by using the $LCD$. The $LCD$ of the denominators $
3
$ and $
6
$ is $
6
$ since it is the lowest number that can be exactly divided by the denominators. Multiplying the fractions by an expression equal to $1$ that will make the denominator equal to the $LCD$ results to
\begin{array}{l}\require{cancel}
\dfrac{2}{3}\cdot\dfrac{2}{2}-\dfrac{5}{6}
\\\\=
\dfrac{4}{6}-\dfrac{5}{6}
\\\\=
\dfrac{4-5}{6}
\\\\=
\dfrac{-1}{6}
\\\\=
-\dfrac{1}{6}
.\end{array}
The expression, $
a^{\frac{3}{4}+\frac{5}{8}}b^{\frac{2}{3}-\frac{5}{6}}
,$ simplifies to
\begin{array}{l}\require{cancel}
a^{\frac{11}{8}}b^{-\frac{1}{6}}
.\end{array}
Using the Negative Exponent Rule of the laws of exponents which states that $x^{-m}=\dfrac{1}{x^m}$ or $\dfrac{1}{x^{-m}}=x^m,$ the expression above is equivalent to
\begin{array}{l}\require{cancel}
\dfrac{a^{\frac{11}{8}}}{b^{\frac{1}{6}}}
\\\\=
\dfrac{a^{11/8}}{b^{1/6}}
.\end{array}
