91Ó°ÊÓ

We have given,

(a)$\frac{a^{\frac{3}{4}}· a^{-\frac{1}{4}}}{a^{-\frac{10}{4}}}\left(b\right){\left(\frac{27b^{\frac{2}{3}}{c}^{-\frac{5}{2}}}{b^{-\frac{7}{3}}{c}^{\frac{1}{2}}}\right)}^{\frac{1}{3}}$

We will use,

Product rule of the exponent,

$ a^b· a^c=a^{b+c}$

Quotient rule of the exponent:

$\frac{x^a}{x^b}=x^{a-b}$

Negative exponent rule,

$a^{-b}=\frac{1}{a^b}$

We have,

$\frac{a^{\frac{3}{4}}· a^{-\frac{1}{4}}}{a^{-\frac{10}{4}}} $

Using product rule,

$\frac{a^{\frac{3}{4}}· a^{-\frac{1}{4}}}{a^{-\frac{10}{4}}} =\frac{a^{\left(\frac{3}{4}+\left(-\frac{1}{4}\right)\right)}}{a^{-\frac{10}{4}}} $

Simplifying,

$\frac{a^{\frac{3}{4}}· a^{-\frac{1}{4}}}{a^{-\frac{10}{4}}} =\frac{a^{\frac{1}{2}}}{a^{-\frac{10}{4}}} $

Using quotient rule of the exponent,

$\frac{a^{\frac{3}{4}}· a^{-\frac{1}{4}}}{a^{-\frac{10}{4}}} =a\left({}^{\frac{1}{2}+\frac{10}{4}}\right) $

$⇒\frac{a^{\frac{3}{4}}· a^{-\frac{1}{4}}}{a^{-\frac{10}{4}}} =a\left({}^{\frac{2}{4}+\frac{10}{4}}\right) $

$⇒\frac{a^{\frac{3}{4}}· a^{-\frac{1}{4}}}{a^{-\frac{10}{4}}} =a^3 $

Hence, value of$\frac{a^{\frac{3}{4}}· a^{-\frac{1}{4}}}{a^{-\frac{10}{4}}} $is$a^3 $.

We have,

${\left(\frac{27b^{\frac{2}{3}}{c}^{-\frac{5}{2}}}{b^{-\frac{7}{3}}{c}^{\frac{1}{2}}}\right)}^{\frac{1}{3}}$

Using Quotient rule,

${\left(\frac{27b^{\frac{2}{3}}{c}^{-\frac{5}{2}}}{b^{-\frac{7}{3}}{c}^{\frac{1}{2}}}\right)}^{\frac{1}{3}}={\left(27b^{\left(\frac{2}{3}+\frac{7}{3}\right)}c\left({}^{-\frac{5}{2}-\frac{1}{2}}\right)\right)}^{\frac{1}{3}}$

$⇒{\left(\frac{27b^{\frac{2}{3}}{c}^{-\frac{5}{2}}}{b^{-\frac{7}{3}}{c}^{\frac{1}{2}}}\right)}^{\frac{1}{3}}={\left(27b^3{c}^{-3}\right)}^{\frac{1}{3}}$

Using power rule of the exponent,

role="math" localid="1645080048276" ${\left(\frac{27b^{\frac{2}{3}}{c}^{-\frac{5}{2}}}{b^{-\frac{7}{3}}{c}^{\frac{1}{2}}}\right)}^{\frac{1}{3}}={\left(27\right)}^{\frac{1}{3}}{b}^{3·\frac{1}{3}}{c}^{-3·\frac{1}{3}}$

role="math" localid="1645080058623" $⇒{\left(\frac{27b^{\frac{2}{3}}{c}^{-\frac{5}{2}}}{b^{-\frac{7}{3}}{c}^{\frac{1}{2}}}\right)}^{\frac{1}{3}}=3b{c}^{-1}$

Using negative exponent rule,

$⇒{\left(\frac{27b^{\frac{2}{3}}{c}^{-\frac{5}{2}}}{b^{-\frac{7}{3}}{c}^{\frac{1}{2}}}\right)}^{\frac{1}{3}}=\frac{3b}{c}$

Hence, value of${\left(\frac{27b^{\frac{2}{3}}{c}^{-\frac{5}{2}}}{b^{-\frac{7}{3}}{c}^{\frac{1}{2}}}\right)}^{\frac{1}{3}}$is$\frac{3b}{c}$.