higher roots problem? equation in replies

Question

anonymous · Answer

$$\sqrt[3]{x^7}$$

anonymous · Answer

Answers: 

$$x *\sqrt[3]{x^2}$$
$$x *\sqrt[3]{x}$$
$$x^3 * 3\sqrt x^2$$
$$x^2 * 3\sqrt x$$

abdullahm · Answer

Let me do an example for you!

$\bf\Large \sqrt[4]{x^{13}} = \sqrt[4]{x^4 \times x^4 \times x^4 \times x}$

$\bf\Large = \sqrt[4]{x^4} \times  \sqrt[4]{x^4} \times  \sqrt[4]{x^4} \times  \sqrt[4]{x}$

$\bf\Large = x \times x \times x \times \sqrt[4]{x}$

$\bf\Large = \boxed{\bf{ x^3\sqrt[4]{x}}}$

abdullahm · Answer

So, basically, you need to factor the term inside in groups of what the nth root is. In my example, it was 4. In your question, it is 3.

Then you can take out as many of the groups as you can until you can't take anymore out.

anonymous · Answer

Where'd the ^13 go? Is it x^4 13 times, and then simplified? o.O D:

abdullahm · Answer

Basically,

$\bf\Large a^b \times a^c = a^{b+c}$

So,

$\bf\Large x^4 \times x^4 \times x^4 \times x = x^{4+4+4+1} = x^{13}$

mathstudent55 · Answer

$\large x^{13} = x^4 \times x^4 \times x^4 \times x = x^{4 + 4 + 4  + 1} = x^{13}$

mathstudent55 · Answer

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