Question

Please help ^3√x^8

Answer 1

@zepdrix

Answer 2

So, hopefully you can get a grasp on this concept,
when the exponent matches the degree of the root,
then we get a full x coming out of the root.

Examples:
\(\large\rm \sqrt[3]{x^3}=x\)
\(\large\rm \sqrt[5]{x^5}=x\)
If our exponent is LARGER than 3,
then we want to break up the exponent into 3's using exponent rules.

Example:
\(\large\rm \sqrt[3]{x^6}=\sqrt[3]{x^3\cdot x^3}=\sqrt[3]{x^3}\cdot\sqrt[3]{x^3}=x\cdot x\)

Answer 3

So for our problem here,
we have 8 x's inside of the root.
So we can break them into a group of 3, 3 and 2, ya?

Answer 4

\[\large\rm \sqrt[3]{x^8}=\sqrt[3]{x^3\cdot x^3\cdot x^2}\]

Answer 5

yes

Answer 6

\(\large\rm =\sqrt[3]{x^3}\cdot\sqrt[3]{x^3}\cdot\sqrt[3]{x^2}\)

Answer 7

So those first two should turn into normal x's, no root, ya?

Answer 8

Okay

Answer 9

\(\large\rm =x\cdot x\cdot\sqrt[3]{x^2}\)
We can group the x's outside of the root,
\(\large\rm =x^2\cdot \sqrt[3]{x^2}\)

Answer 10

That one way to simplify the expression.
You didn't provide any instructions though,
so I dunno if that's what you were looking for or not.

Answer 11

also, \(\sqrt[a]{b^c}=b^{\frac{c}{a}}\)

Answer 12

it can also be x8/3 right?

Answer 13

yes :) x^(8/3)