Bginner

The statement you're trying to prove is obviously false and it's false where you said you checked.  Take n = 0.  Then $5^{n^3}+1 = 5^0+1 = 2$.  Also, $3^{n+1} = 3^1 = 3$.  Obviously 2 is not divisible by 3.
 

This question has been asked before.  See a solution at https://web2.0calc.com/questions/polynomial_80#r1.  You may also want to see https://web2.0calc.com/questions/hard-algebra_17#r1 which solves this kind of problem but it has more explicit details.   If you don't understand some steps, ask.

This question has been asked before.   See a solution at https://web2.0calc.com/questions/hard-algebra_17#r1

Your idea of using the inequality on AM and GM is correct.

Let $\alpha, \beta, \gamma$ be the real nonnegative roots of the given polynomial,
thus, $x^3+ax^2+bx+c = (x-\alpha)(x-\beta)(x-\gamma)$.  By Vieta's formula,
$\alpha\beta+\beta\gamma+\gamma\alpha = b$ and $\alpha\beta\gamma=c$.

Can you do the rest?

Note that $1/(abc) + 1/(abd) + 1/(acd) + 1/(bcd) = (a+b+c+d)/(abcd)$.  The values of $a+b+c+d$ and $abcd$ can be read off from the coefficients of $x^4-7x^3+4x^2+7x-4=0$ by Vieta's formula.
 

Completing the square on $x$ and $y$ independently to write the given expression as
$(x+a)^2+(y+b)^2+c$.   The answer is $c$.
 

If you make a table of congruences modulo 6 of

$$5^n, 8^{n+1}, 13^{n+2}, 14^{n+3)$$

for $n \equiv 0,1,2,3,4,5\pmod6$, you'll see a pattern.  According to my table, half of $$5^n + 8^{n+1} + 13^{n+2} + 14^{n+3)$$ for $1\le n< 100$ are $\equiv 0\pmod6$ and the other half are $\equiv 2\pmod6$.

For conveniece, when creating the table, replace every base $a$ by $a \bmod 6$.
 

Excuse me for not understanding.  If the sequence consists of all positive multiples of 3 that contain at least one digit that is a 1, shouldn't numbers like 111, 114, 141, 117, 171, etc. belong to the sequence as well?
 

Try plugging in 1/5 for y in the given equation and then try solving for x.

OK then.   You can still use my technique for this problem.
Note that \[\dfrac{x+\dfrac{1}{x}}{x-\dfrac{1}{x}} =
\dfrac{x^2+1}{x^2-1}.\]
So
\begin{eqnarray*}
    \dfrac{x^2+1}{x^2-1} &=& \dfrac{5}{4} \\
    \dfrac{(x^2+1)+(x^2-1)}{(x^2+1)-(x^2-1)} &=& \dfrac{5+4}{5-4} \\
    \dfrac{2x^2}{2} &=& \dfrac{9}{1} \\
    x^2 &=& 9 \\
    x &=& \pm 3.
\end{eqnarray*}

I will avoid expanding any of the cubed expression.  Define a polynomial $p$ to be $p(x)=(x+4)^3 + (x+5)^3 - (x+7)^3 - (x-2)^3$.  Note that $p$ is at most quadratic.  Thus, $p(x)=ax^2+bx+c$ for some real $a,b,c$.  (In fact, $a,b,c$ are integers, but it's enough to say they are real.) We have
\begin{eqnarray*}
    p(0) = c &=& 4^3+5^3-7^3 - (-2)^3 = -146 \\
    p(1) = a+b+c &=& 5^3+6^3-8^3-(-1)^3 = -170 \\
    p(-1) = a-b+c &=& 3^3+4^3-6^3-(-3)^3 = -98
\end{eqnarray*}
Solving for $a$, $b$, and $c$, we have $a=12$, $b=-36$, and $c=-146$, giving $p(x)= 12x^2-36x-146$.   So
\[
    x = \dfrac{-b\pm\sqrt{b^2-4ac}}{2a} = \dfrac{3}{2}\pm\dfrac{1}{6\sqrt{519}}.
\]

Please delete that.   I read the question wrong.  Sorry.

There's nothing wrong with the theorem I showed to you.   Well, I proved it, didn't I :)
So it could be wrong only if my proof were invalid!

Now let me explain how I concluded that this problem could be solved by simply
taking the HM of the two given speeds.   

Let x be the time it takes in minutes to arrive on time.
To arrive five minutes late means the journey takes x+5 minutes.
To arrive five minutes early means the journey takes x-5 minutes.

We are given that if r1 = 45, t1 = x+5 and
if r2 = 63, t2 = x-5.`

So ta = ((x+5)+(x-5))/2 = x minutes, exactly the the AM of t1 and t2.
(It's immaterial whether you convert the times to hour or not.
Were t1 and t2 to be counted in hours, its AM would still be the on-time time.
Why?  Because the late amount and the early amount equals, so they cancel out
when you calculate the AM.  The problem poser intended it that way.)

So according to the theorem, ra should be the HM of r1 and r2.

Now back to the case of n = 3.
Were there to be a distance d, and speeds r1, r2, r3, and desired speed x
such that (t1 + t2 + t3) / 3 = x, then the AM of r1, r2, r3 would be x and
ra, the HM of the three speeds, would be the answer as well.
However, these numbers r1, r2, r3, t1, t2, t3 must have come from reality.
In a hurry I simply picked arbitrary numbers and used them in my example.
That's bad.  These numbers aren't arbitrary.  They're dependent on one another.
So that's why you get funny result.