Answer:
![\sqrt[3]{3}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B3%7D)
Step-by-step explanation:
Our expression is: ![\frac{1}{3} \sqrt[3]{81}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B3%7D%20%5Csqrt%5B3%5D%7B81%7D)
.
Let's focus on the cube root of 81 first. What's the prime factorisation of 81? It's simply: 3 * 3 * 3 * 3, or 
. Put this in for 81:
![\sqrt[3]{81} =\sqrt[3]{3^3*3}=\sqrt[3]{3^3} *\sqrt[3]{3}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B81%7D%20%3D%5Csqrt%5B3%5D%7B3%5E3%2A3%7D%3D%5Csqrt%5B3%5D%7B3%5E3%7D%20%2A%5Csqrt%5B3%5D%7B3%7D)
We know that the cube root of 3 cubed will cancel out to become 3, but the cube root of 3 cannot be further simplified, so we keep that. Our outcome is then:
![\sqrt[3]{3^3} *\sqrt[3]{3}=3\sqrt[3]{3}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B3%5E3%7D%20%2A%5Csqrt%5B3%5D%7B3%7D%3D3%5Csqrt%5B3%5D%7B3%7D)
Now, let's multiply this by 1/3, as shown in the original problem:
![\frac{1}{3}* 3\sqrt[3]{3}=\sqrt[3]{3}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B3%7D%2A%203%5Csqrt%5B3%5D%7B3%7D%3D%5Csqrt%5B3%5D%7B3%7D)
Thus, the answer is .
<em>~ an aesthetics lover</em>
About 145 I gonna guess I dont know if that's right
Answer: 3
Step-by-step explanation:
The question is to calculate one-eight of the sum of 18 and the product of 6 and 1. This will be:
= 1/8 [18 + (6 × 1)]
= 1/8 [18 + 6]
= 1/8 × 24
= 3
The answer is 3.
No it’s not a perfect cube
<span><span>(<span><span>(<span>−3</span>)</span><span>(2)</span></span>)</span><span>(3)</span></span><span>=<span><span>(<span>−6</span>)</span><span>(3)</span></span></span><span>=<span>−<span>18</span></span></span>
