Calculating the Sixth Root of 3

[mathdad]

Let cbrt = cube rootcbrt{3} x cbrt{3} =

(3)^(1/3) * (3)^(1/3)

3^(1/6) ir sixth root {3}

Correct?

 

[MarkFL]

In general, you want to use the following rule:

\(\displaystyle \sqrt[3]{a}\cdot\sqrt[3]{b}=\sqrt[3]{ab}\)

Using exponents, we can get the same result:

\(\displaystyle a^c\cdot b^c=(ab)^c\)

Now, when the base is the same, we can simply add exponents:

\(\displaystyle a^b\cdot a^c=a^{b+c}\)

So, in the given expression, we may write:

\(\displaystyle \sqrt[3]{3}\cdot\sqrt[3]{3}=\sqrt[3]{3\cdot3}=\sqrt[3]{3^2}=3^{\frac{2}{3}}=3^{\frac{1}{3}+\frac{1}{3}}=3^{\frac{1}{3}}\cdot3^{\frac{1}{3}}=\sqrt[3]{3}\cdot\sqrt[3]{3}\)

 

[mathdad]

In general, you want to use the following rule:

\(\displaystyle \sqrt[3]{a}\cdot\sqrt[3]{b}=\sqrt[3]{ab}\)

Using exponents, we can get the same result:

\(\displaystyle a^c\cdot b^c=(ab)^c\)

Now, when the base is the same, we can simply add exponents:

\(\displaystyle a^b\cdot a^c=a^{b+c}\)

So, in the given expression, we may write:

\(\displaystyle \sqrt[3]{3}\cdot\sqrt[3]{3}=\sqrt[3]{3\cdot3}=\sqrt[3]{3^2}=3^{\frac{2}{3}}=3^{\frac{1}{3}+\frac{1}{3}}=3^{\frac{1}{3}}\cdot3^{\frac{1}{3}}=\sqrt[3]{3}\cdot\sqrt[3]{3}\)

Great but is my answer wrong?

 

[MarkFL]

Great but is my answer wrong?

Yes, your result is incorrect. :D

You want to add the two exponents to get 1/3 + 1/3 = 2/3.

 

[mathdad]

Yes, your result is incorrect. :D

You want to add the two exponents to get 1/3 + 1/3 = 2/3.

I forgot that powers are added.

 

[HallsofIvy]

Well, what in the world did you do to get "1/6"?

 

[mathdad]

Thank you everyone.