Answer:

Step-by-step explanation:
<u>Roots of a polynomial</u>
If we know the roots of a polynomial, say x1,x2,x3,...,xn, we can construct the polynomial using the formula

Where a is an arbitrary constant.
We know three of the roots of the degree-5 polynomial are:

We can complete the two remaining roots by knowing the complex roots in a polynomial with real coefficients, always come paired with their conjugates. This means that the fourth and fifth roots are:

Let's build up the polynomial, assuming a=1:

Since:


Operating the last two factors:

Operating, we have the required polynomial:

$6 for each popcorn and $4 for each soda
Sum of the terms of the series is
Sn = n/2 ( a1+an )
we have n= 6 , a1= 17, an = 57
so Sn = 6/2 ( 17+57) = 3(74) = 222
Answer:
The missing value is 512
![8=\sqrt[3]{512}](https://tex.z-dn.net/?f=8%3D%5Csqrt%5B3%5D%7B512%7D)
Step-by-step explanation:
Let
x ----> the missing value
we have
![8=\sqrt[3]{x}](https://tex.z-dn.net/?f=8%3D%5Csqrt%5B3%5D%7Bx%7D)
solve for x
That means-----> isolate the variable x
Elevated both sides to the cube
![8^{3}=(\sqrt[3]{x})^{3}](https://tex.z-dn.net/?f=8%5E%7B3%7D%3D%28%5Csqrt%5B3%5D%7Bx%7D%29%5E%7B3%7D)
Remember that
![(\sqrt[3]{a})^{3}=(a^{\frac{1}{3}})^3=a](https://tex.z-dn.net/?f=%28%5Csqrt%5B3%5D%7Ba%7D%29%5E%7B3%7D%3D%28a%5E%7B%5Cfrac%7B1%7D%7B3%7D%7D%29%5E3%3Da)
so
![(\sqrt[3]{x})^{3}=x](https://tex.z-dn.net/?f=%28%5Csqrt%5B3%5D%7Bx%7D%29%5E%7B3%7D%3Dx)
substitute


Rewrite

so
the equation is
![8=\sqrt[3]{512}](https://tex.z-dn.net/?f=8%3D%5Csqrt%5B3%5D%7B512%7D)