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How do you simplify $\sqrt[3]{{\dfrac{{16}}{{27}}}}$?

Answer
VerifiedVerified
528.9k+ views
Hint: First, write the numerator and denominator separately in the given fraction. Then, simplify the numerator by rewriting $16$ as ${2^3} \cdot 2$, then factor 8 out of 16, then pull terms out from under the radical. Then, simplify the denominator by rewriting 27 as ${3^3}$ pulling terms out from under the radical, assuming real numbers. We will get the simplified version of $\sqrt[3]{{\dfrac{{16}}{{27}}}}$.
Formula used:
i). Power rule to combine exponents: ${a^m} \times {a^n} = {a^{m + n}}$
ii). $\sqrt[n]{{{a^x}}} = {a^{\dfrac{x}{n}}}$
iii). ${\left( {{a^m}} \right)^n} = {a^{mn}}$

Complete step-by-step solution:
We have to simplify $\sqrt {\dfrac{5}{3}} $.
So, first rewrite $\sqrt[3]{{\dfrac{{16}}{{27}}}}$ as $\dfrac{{\sqrt[3]{{16}}}}{{\sqrt[3]{{27}}}}$.
$ \Rightarrow \dfrac{{\sqrt[3]{{16}}}}{{\sqrt[3]{{27}}}}$
Now, simplify the numerator.
Rewrite $16$ as ${2^3} \cdot 2$.
We can factor 8 out of 16, in above numerator
$ \Rightarrow \dfrac{{\sqrt[3]{{8 \times 2}}}}{{\sqrt[3]{{27}}}}$
We can rewrite 8 as ${2^3}$ in above numerator
$ \Rightarrow \dfrac{{\sqrt[3]{{{2^3} \times 2}}}}{{\sqrt[3]{{27}}}}$
Pull terms out from under the radical by using property that ${a^{m \times \dfrac{1}{m}}} = a$.
$ \Rightarrow \dfrac{{2\sqrt[3]{2}}}{{\sqrt[3]{{27}}}}$
Now, simplify the denominator.
Rewrite 27 as ${3^3}$.
$ \Rightarrow \dfrac{{2\sqrt[3]{2}}}{{\sqrt[3]{{{3^3}}}}}$
Pull terms out from under the radical, assuming real numbers by using property that ${a^{m \times \dfrac{1}{m}}} = a$.
$ \Rightarrow \dfrac{{2\sqrt[3]{2}}}{3}$
The result can be shown in multiple forms.
Exact Form: $\dfrac{{2\sqrt[3]{2}}}{3}$
Decimal Form: $0.8399473666$
Hence, simplified version of $\sqrt[3]{{\dfrac{{16}}{{27}}}}$ is $\dfrac{{2\sqrt[3]{2}}}{3}$.

Note: By simplifying a fraction, we mean to express the fraction as a ratio of prime numbers or we can say that both the numerator and denominator should be prime numbers, that is, they should be divisible by only $1$ and itself. For simplifying a fraction, we write it as a product of prime factors, and then divide both of them with the common factors, in this question both the numerator and denominator are already prime numbers and thus the fraction cannot be simplified further.

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