Logan Glover

2023-03-13

What is the cube root of $27{a}^{12}$?

kuskdtq

Beginner2023-03-14Added 4 answers

Assign the term we are looking for the number n. We can then express this issue as follows:

$n=\sqrt[3]{27{a}^{12}}$

And, because $\sqrt[{n}]{x}={x}^{\frac{1}{{n}}}$ we can then rewrite it as:

$n={\left(27{a}^{12}\right)}^{\frac{1}{3}}$

Next, we can rewrite 27 as:

$n={\left({3}^{3}{a}^{12}\right)}^{\frac{1}{3}}$

Now, we can use the rule of exponents to eliminate the exponent outside the parenthesis: $\left({x}^{{a}}\right)}^{{b}}={x}^{{a}\times {b}$

$n={\left({3}^{{3}}{a}^{{12}}\right)}^{{\frac{1}{3}}}$

$n={3}^{{3}\times {\frac{1}{3}}}{a}^{{12}\times {\frac{1}{3}}}$

$n={3}^{\frac{3}{3}}{a}^{\frac{12}{3}}$

$n={3}^{1}{a}^{4}$

And by applying this exponents rule, we can finish the equation:

${a}^{{1}}=a$

$n={3}^{{1}}{a}^{4}$

$n=3{a}^{4}$

$n=\sqrt[3]{27{a}^{12}}$

And, because $\sqrt[{n}]{x}={x}^{\frac{1}{{n}}}$ we can then rewrite it as:

$n={\left(27{a}^{12}\right)}^{\frac{1}{3}}$

Next, we can rewrite 27 as:

$n={\left({3}^{3}{a}^{12}\right)}^{\frac{1}{3}}$

Now, we can use the rule of exponents to eliminate the exponent outside the parenthesis: $\left({x}^{{a}}\right)}^{{b}}={x}^{{a}\times {b}$

$n={\left({3}^{{3}}{a}^{{12}}\right)}^{{\frac{1}{3}}}$

$n={3}^{{3}\times {\frac{1}{3}}}{a}^{{12}\times {\frac{1}{3}}}$

$n={3}^{\frac{3}{3}}{a}^{\frac{12}{3}}$

$n={3}^{1}{a}^{4}$

And by applying this exponents rule, we can finish the equation:

${a}^{{1}}=a$

$n={3}^{{1}}{a}^{4}$

$n=3{a}^{4}$