In the following exercises, simplify.

$ⓐ{25}^{\frac{3}{2}}ⓑ9^{-\frac{3}{2}}ⓒ{(-64)}^{\frac{2}{3}}$

We have given,

$ⓐ{25}^{\frac{3}{2}}ⓑ9^{-\frac{3}{2}}ⓒ{(-64)}^{\frac{2}{3}}$

Here we are going to use,

Negative exponent rule:

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

Power rule of exponent:

${\left(a^b\right)}^c=a^{bc}$

We have,

${25}^{\frac{3}{2}}$

We will find the factors of 25,

$25=5\times 5=5^2$

Thus,

role="math" localid="1645014424162" ${25}^{\frac{3}{2}}={\left(5^2\right)}^{\frac{3}{2}}$

Using power rule of exponent,

${25}^{\frac{3}{2}}={\left(5\right)}^3$

$5^3=5\times 5\times 5=125$

role="math" localid="1645014479619" $\Rightarrow {25}^{\frac{3}{2}}=125$

Hence, value of the${25}^{\frac{3}{2}}$is 125.

We have given,

$9^{-\frac{3}{2}}$

Using negative exponent rule,

$9^{-\frac{3}{2}}=\frac{1}{9^{\frac{3}{2}}}$

We will find the factors of 9,

$9=3\times 3=3^2$

Thus,

$9^{-\frac{3}{2}}=\frac{1}{{\left(3^2\right)}^{\frac{3}{2}}}$

Using power rule of the exponent,

$9^{-\frac{3}{2}}=\frac{1}{3^3}$

$3^3=3\times 3\times 3=27$

$\Rightarrow 9^{-\frac{3}{2}}=\frac{1}{27}$

Hence, value of $9^{-\frac{3}{2}}$is$\frac{1}{27}$.

We have given,

${(-64)}^{\frac{2}{3}}$

We will find the factors of 64,

$64=4\times 4\times 4$

Thus,

${(-64)}^{\frac{2}{3}}={(-4^3)}^{\frac{2}{3}}$

Using power rule of the exponent,

${(-64)}^{\frac{2}{3}}={(-4)}^2$

${(-4)}^2=(-4)(-4)=16$

$\Rightarrow {(-64)}^{\frac{2}{3}}=16$

Hence, value of${(-64)}^{\frac{2}{3}}$is 16.