Voldemort

Guest, could you explain how you got that answer?
 

(x-a) + (x - b) + (x-c) = 3x - (a + b + c) = 0

a + b + c = 4 by vietas

3x = 4

x = 4/3.

5^n + 8^(n+1) + 13^(n + 2)  = (-1)^n + 2^(n+1) + 1  (mod 6)

Case 1: n is odd

2^(n + 1) = 0 (mod 6)

No values of n will satisfy this as we will need to get a factor of 3 in order to make this expression divisible by 6. 

Case 2: n is even

2^(n+1) + 2 = 0 (mod 6)

2^(n+1) = 4 (mod 6)

2

4 check

8 

16 check

32 
64 check
128

256 check

It seems like all the even powers of 2 leave a remainder of 4 when divided by 6.

However, if we subtract one, we would get an odd number for n which goes against the case restrictions. Therefore, there are 0?

Please correct me if I am wrong.

Use the angle bisector theorem! Set AZ = x and BZ = 8 - x. Can you use this theorem from here?

There are actually 6 stars and 5 bars, as there are 6 of these integers. You should be able to sovle the problem now, Builderboi just made a tiny error. 

We can remove 1/x as 1/x would approach 0 when x is approaching infinity.

For any number x, 1^x = 1. So the limit does exist :)

f(x) = (x - r)(x - s)(x - t)(x - u)

f(-2) = (-2 - r)(-2 - s)(-2 - t)(-2-u) = (2 + r)(2 + s)(2 + t)(2 + u).

Solve for f(-2) and you will receive your answer!

What is it specifically asking us to find?

Use the law of cosines, and find the area using Heron's formula.

I'll use Sophie Germain's Identity for this problem and don't think it would be posssible to use the Simon's favorite factoring trick. The Simon's favorite factoring trick is only applicable in:

- Two variable problems

- Degree of no more than 2. 

z^4 + 4(1^4) = (z^2 + 2z + 2)(z^2 - 2z + 2)

You can find the complex roots by using the quadratic formula from there.

For this problem, you can even use Roots of Unity: https://artofproblemsolving.com/wiki/index.php/Roots_of_unity.

Use the Cauchy Schwarz inequality!