Logan Glover

2023-03-13

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

### Answer & Explanation

kuskdtq

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: ${\left({x}^{a}\right)}^{b}={x}^{a×b}$
$n={\left({3}^{3}{a}^{12}\right)}^{\frac{1}{3}}$
$n={3}^{3×\frac{1}{3}}{a}^{12×\frac{1}{3}}$
$n={3}^{\frac{3}{3}}{a}^{\frac{12}{3}}$
$n={3}^{1}{a}^{4}$
And by applying this exponents rule, we can finish the equation:
${a}^{1}=a$
$n={3}^{1}{a}^{4}$
$n=3{a}^{4}$

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