In the following exercises, simplify.

$ⓐ{625}^{\frac{1}{4}}ⓑ{243}^{\frac{1}{5}}ⓒ{32}^{\frac{1}{5}}$

Given fractional numbers are $ⓐ{625}^{\frac{1}{4}}ⓑ{243}^{\frac{1}{5}}ⓒ{32}^{\frac{1}{5}}$

If $\sqrt[n]{a}$is a real number and role="math" localid="1644910916649" $n\ge 2$then $\sqrt[n]{a}=a^{\frac{1}{n}}$.

Also,

${\left(a\right)}^{\frac{n}{n}}=a$

Powers can be written in the multiplication like as,$a^n=a\times a\times a\times a\times ...\times a(ntimes)$.

Power property:${\left(a^m\right)}^n=a^{m·n}$

We have,

${625}^{\frac{1}{4}}$

We will find the prime factors of 625.

role="math" localid="1644911248266" $625=5\times 5\times 5\times 5=5^4$

Thus,

${625}^{\frac{1}{4}}={\left(5^4\right)}^{\frac{1}{4}}$

Using power property,

${625}^{\frac{1}{4}}=5^{4·\frac{1}{4}}=5$

Hence,$\frac{1}{6}$th power of 625 is 5.

We have,

${243}^{\frac{1}{5}}$

We will find the prime factors of 243,

$243=3\times 3\times 3\times 3\times 3$

Rewriting factors in to the power form,

$243=3^5$

Thus,

${243}^{\frac{1}{5}}={\left(3^5\right)}^{\frac{1}{5}}$

Using power property,

${243}^{\frac{1}{5}}=3$

Hence, $\frac{1}{5}$th power of 243 is 3.

We have,

${32}^{\frac{1}{5}}$

We will find the prime factors of 32,

$32=2\times 2\times 2\times 2\times 2$

Rewriting it in to the power form,

$32=2^5$

Thus,

${32}^{\frac{1}{5}}={\left(2^5\right)}^{\frac{1}{5}}$

Using power rule,

${32}^{\frac{1}{5}}=2$

Hence,$\frac{1}{5}$th power of 32 is 2.