Secondary school level mathematical induction 
*

*It is given that 
$$1^3+2^3+3^3+\cdots+n^3=\frac{n^2(n+1)^2}{4}$$


Then, how to find the value of 
$2^3+4^3+\cdots+30^3$? 
Which direction should I aim at?


*Prove by mathematical induction, that $5^n-4^n$ is divisible by 9 for all positive even numbers $n$.


$$5^n-4^n=9m,\text{where $m$ is an integer.}$$
What I am thinking in the $n+1$ step is,
\begin{align}
& 5^{n+2}-4^{n+2} \\
= {} & 5^2(5^n-4^n)+5^24^n-4^{n+2} \\
= {} & 5^2(5^n-4^n)+4^n(5^2-4^2) \\
= {} & 5^29m+4^n9 \\
= {} & 9(5^2m+4^n)
\end{align}
Does this approach make sense?


*Show that $a+b$ is a factor of $a^n+b^n$ where $n$ is a positive odd number.


I am thinking this in the $n+1$ step.
$$a^{2n+1}+b^{2n+1}$$
But then I cannot get it further.

Hints
1) $2^3+4^3+...+30^3=2^3(1^3+2^3+...+15^3)$ now apply the formula you got.
2) Assume that $9k=5^n-4^n$ then $5^{n+2}-4^{n+2}=25.5^n-16.4^n=25(9k+4^n)-16.4^n=9k'+9.4^n$
3) Change the questions to show that $(a+b)|(a^{2k+1}+b^{2k+1})$ for all $k\in\mathbb{N}$.Let $(a+b)x=a^{2k+1}+b^{2k+1}$ now $a^{2k+3}+b^{2k+3}=a^2.a^{2k+1}+b^2.b^{2k+1}$ again do substitution and it will work out.
