Question

Assume all vars are positive reals​

Answer 1

Answer:

$\frac{x^{3}\sqrt[3]{x} }{4}$

Step-by-step explanation:

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What this is saying is it is taking the cubed root (which is like a square root, except it is using a 3 instead of 2) of x to the power of 10 divided by 64.

The equation is also saying:

$\frac{\sqrt[3]{x^{10} }}{\sqrt[3]{64} }$

We can simplify the bottom half of the fraction to be 4. The cube root of something is essentially saying its the number times itself 3 times (in this case, divided by that original number). The number to get to 64 was 4 and it was done like this: 4 times 4 times 4 (which equals 64).

As for the top equation, it will simplify down to $x^{3} \sqrt[3]{x}$.

Since $x^{10}$ can be broken up to be x and $x^{9}$, we cube root the $x^{9}$ to become $x^{3}$, while the x remains the same.

We come to result of $\frac{x^{3}\sqrt[3]{x} }{4}$, which is the answer.

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