Question

Square root of 2 divided by the cube root of 2

Answer 1

ANSWER

$\frac{ \sqrt{2} }{ \sqrt[3]{2} } = \sqrt[6]{2}$

EXPLANATION

We want to simplify

$\frac{ \sqrt{2} }{ \sqrt[3]{2} }$

We rewrite in exponential form to get:

$\frac{ \sqrt{2} }{ \sqrt[3]{2} } = \frac{ {2}^{ \frac{1}{2} } }{{2}^{ \frac{1}{3} } }$

Recall the quotient rule of exponents:

$\frac{ {a}^{m} }{ {a}^{n} } = {a}^{m - n}$

We apply this rule to get:

$\frac{ \sqrt{2} }{ \sqrt[3]{2} } = {2}^{ \frac{1}{2} - \frac{1}{3} }$

Simplify:

$\frac{ \sqrt{2} }{ \sqrt[3]{2} } = {2}^{ \frac{3 - 2}{6} }$

$\frac{ \sqrt{2} }{ \sqrt[3]{2} } = {2}^{ \frac{1}{6} }$

$\frac{ \sqrt{2} }{ \sqrt[3]{2} } = \sqrt[6]{2}$

Answer 2

Answer: $\sqrt[6]{2}$

Step-by-step explanation:

You know that the expression is $\frac{\sqrt{2}}{\sqrt[3]{2}}$

By definition we know that:

$\sqrt[n]{a}=a^{\frac{1}{n}$

You also need to remember the Quotient of powers property:

$\frac{a^n}{a^m}=a^{(n-m)}$

Therefore, you can rewrite the expression:

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

Finally, you have to simplify the expression. Therefore, you get:

$=2^{(\frac{1}{2}-\frac{1}{3})}\\=2^{\frac{1}{6}}\\=\sqrt[6]{2}$