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You wish to prove $$(1+2+\dots+n)^2+(n+1)^3 = (1+2+\dots+(n+1))^2.$$
To get it to look similar, expand the RHS like we do this: $(a+b)^2 = a^2+2ab + b^2$.
$$ ( 1+ 2 + \cdots + (n+1) )^2 = (1+2+\cdots + n)^2 + 2\cdot (1+2+\cdots +n) (n+1) + (n+1)^2 .$$
Use the fact that $ 1+2 + 3 + \cdots + n = \frac{ n(n+1)}{2} $ so the term on the RHS simplifies to
$$ (1+2+\cdots + n)^2 + n(n+1)^2 + (n+1)^2 = (1+2+\cdots + n)^2 + (n+1)^2(n+1) $$
$$ = (1+2+\cdots + n)^2 + (n+1)^3 $$
as required.
For the second question, yes the method of induction works well on that example, try it!
In response to Andrew's comment:
I think you would need to use that fact in most induction proofs of this statement. But you needn't view it as an advanced fact we had to bring in, it's quite simple. Indeed, write $ S = 1+2+3+\cdots + (n-1) + n $ and then right underneath that line, write the sum in reverse order: $ S = n + (n-1) + \cdots + 3 + 2 + 1 $. So if we add these we get $ 2S = (n+1) + (n+1) + \cdots + (n+1) + (n+1) + (n+1) $, where there are $n$ terms on the right hand side. The result follows. Or, since you are practicing, a proof by induction works as well!
For your second issue: The important part is the idea, then you will see you don't have to do much differently for different types of questions.
Outline: i) If $P(k)$ is any statement that depends on a natural number $k$, such that $P(k)$ being true implies $ P(k+1) $ is true, then if we check $P(1) $ is true, it is true for all integers. Why?
We check $P(1) $ is true manually. But this implies $P(1) $ is true, which in turn implies $P(2) $, \cdots....
These statements don't need to be equalities, they can be inequalities, or divisibility properties, or many many other things.
