Answer:
The third option listed: ![\sqrt[3]{2x} -6\sqrt[3]{x}\\](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B2x%7D%20-6%5Csqrt%5B3%5D%7Bx%7D%5C%5C)
Step-by-step explanation:
We start by writing all the numerical factors inside the qubic roots in factor form (and if possible with exponent 3 so as to easily identify what can be extracted from the root):
![7\sqrt[3]{2x} -3\sqrt[3]{16x} -3\sqrt[3]{8x} =\\=7\sqrt[3]{2x} -3\sqrt[3]{2^32x} -3\sqrt[3]{2^3x} =\\=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%5Csqrt%5B3%5D%7B2x%7D%20%20-3%5Csqrt%5B3%5D%7B16x%7D%20-3%5Csqrt%5B3%5D%7B8x%7D%20%3D%5C%5C%3D7%5Csqrt%5B3%5D%7B2x%7D%20%20-3%5Csqrt%5B3%5D%7B2%5E32x%7D%20-3%5Csqrt%5B3%5D%7B2%5E3x%7D%20%3D%5C%5C%3D7%5Csqrt%5B3%5D%7B2x%7D%20%20-3%2A2%5Csqrt%5B3%5D%7B2x%7D%20-3%2A2%5Csqrt%5B3%5D%7Bx%7D%3D%5C%5C%3D7%5Csqrt%5B3%5D%7B2x%7D%20%20-6%5Csqrt%5B3%5D%7B2x%7D%20-6%5Csqrt%5B3%5D%7Bx%7D)
And now we combine all like terms (notice that the only two terms we can combine are the first two, which contain the exact same radical form:
![7\sqrt[3]{2x} -6\sqrt[3]{2x} -6\sqrt[3]{x}=\\=\sqrt[3]{2x} -6\sqrt[3]{x}](https://tex.z-dn.net/?f=7%5Csqrt%5B3%5D%7B2x%7D%20%20-6%5Csqrt%5B3%5D%7B2x%7D%20-6%5Csqrt%5B3%5D%7Bx%7D%3D%5C%5C%3D%5Csqrt%5B3%5D%7B2x%7D%20-6%5Csqrt%5B3%5D%7Bx%7D)
Therefore this is the simplified radical expression: ![\sqrt[3]{2x} -6\sqrt[3]{x}\\](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B2x%7D%20-6%5Csqrt%5B3%5D%7Bx%7D%5C%5C)
<h3>-25.5 < -20.5</h3>
Step-by-step explanation:
<h3>Hope it helps you !!!</h3>
If you multiple 128 by 6 you will get your answer of 768.
Answer:
It should be A
Step-by-step explanation:
Because there are 4 sides and if you multiply 16x4 that will give you 64
it simple good luck :)
Answer:
The number of complex roots is 6.
Step-by-step explanation:
Descartes's rule of signs tells you that the number of positive real roots is 0. The number of negative real roots will be at most 2. The minimum value of the left side will be between x=0 and x=-1, but will never be negative. Thus all six roots are complex.
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The magnitude of x^3 will exceed the magnitude of x^6 only for values of x between -1 and 1. Since the magnitude of either of these terms will not be more than 1 in that range, the left-side expression must be positive everywhere.