For this case we must simplify the following expression:![7 (\sqrt [3] {2x}) - 3 (\sqrt [3] {16x}) - 3 (\sqrt [3] {8x})](https://tex.z-dn.net/?f=7%20%28%5Csqrt%20%5B3%5D%20%7B2x%7D%29%20-%203%20%28%5Csqrt%20%5B3%5D%20%7B16x%7D%29%20-%203%20%28%5Csqrt%20%5B3%5D%20%7B8x%7D%29)
We rewrite:

We rewrite the expression:
![7 (\sqrt [3] {2x}) - 3 (\sqrt [3] {2 ^ 3 * 2x}) - 3 (\sqrt [3] {2 ^ 3 * x}) =](https://tex.z-dn.net/?f=7%20%28%5Csqrt%20%5B3%5D%20%7B2x%7D%29%20-%203%20%28%5Csqrt%20%5B3%5D%20%7B2%20%5E%203%20%2A%202x%7D%29%20-%203%20%28%5Csqrt%20%5B3%5D%20%7B2%20%5E%203%20%2A%20x%7D%29%20%3D)
By definition of properties of powers and roots we have:
![\sqrt [n] {a ^ m} = a ^ {\frac {m} {n}}](https://tex.z-dn.net/?f=%5Csqrt%20%5Bn%5D%20%7Ba%20%5E%20m%7D%20%3D%20a%20%5E%20%7B%5Cfrac%20%7Bm%7D%20%7Bn%7D%7D)
Then, taking the terms of the radical:
![7 (\sqrt [3] {2x}) - 3 (2 \sqrt [3] {2x}) - 3 (2 \sqrt [3] {x}) =\\7 \sqrt [3] {2x} -6 \sqrt [3] {2x} -6 \sqrt [3] {x} =](https://tex.z-dn.net/?f=7%20%28%5Csqrt%20%5B3%5D%20%7B2x%7D%29%20-%203%20%282%20%5Csqrt%20%5B3%5D%20%7B2x%7D%29%20-%203%20%282%20%5Csqrt%20%5B3%5D%20%7Bx%7D%29%20%3D%5C%5C7%20%5Csqrt%20%5B3%5D%20%7B2x%7D%20-6%20%5Csqrt%20%5B3%5D%20%7B2x%7D%20-6%20%5Csqrt%20%5B3%5D%20%7Bx%7D%20%3D)
We add similar terms:
![\sqrt [3] {2x} -6 \sqrt [3] {x}](https://tex.z-dn.net/?f=%5Csqrt%20%5B3%5D%20%7B2x%7D%20-6%20%5Csqrt%20%5B3%5D%20%7Bx%7D)
Answer:
Option C
Answer:
yes
Step-by-step explanation:
If you are dividing it will become 1 because 10 minus 10 equals 0. Anything to the power of 0 is 1 then you keep the base which is y so the answer is y
If you are multipling the answer will be y^10 because 10 plus 10 is 20 and you keep the base which is y