How can you prove $1^3 + 2^3+\cdots+(n-1)^3 < \frac{n^4}{4} < 1^3 + 2^3 + \cdots + n^3$ by induction? Can you provide the steps and corresponding explanations to prove the following predicate by induction?
$$P(n) := 1^3 + 2^3+\cdots+(n-1)^3 < \frac{n^4}{4} < 1^3 + 2^3 + \cdots + n^3$$
I've done some work on it myself by attempting to show that $\frac{k^4}{4} < \frac{(k + 1)^4}{4}$ for the RHS, but I don't understand exactly what I am doing.
Thank you.
Notice: This is not a homework question. I'm attempting to self-study Calculus over the Summer.

Following up on my comment from above, I'm setting out to prove that
$$
\frac{(n-1)^4}4\leq 1+\cdots +(n-1)^3\leq\frac{n^4}4
$$
for suitable $n$ (say $n\geq 1$). (I feel that this ought to be easier as I only have one long sum of cubes to contend with rather than two.)
The base case is easily shown: $0\leq 0\leq \frac14$.
As for the induction step, assume $k\geq 1$ and that
$$
\frac{(k-1)^4}4\leq 1+\cdots +(k-1)^3\leq\frac{k^4}4
$$
Adding $k^3$ to all sides, we get
$$
\frac{(k-1)^4 + 4k^3}4\leq 1+\cdots +(k-1)^3 + k^3\leq\frac{k^4+4k^3}4
$$
Now in the numerator on the right-hand side we have
$$
k^4 + 4k^3\leq k^4 + 4k^3+6k^2+4k+1 = (k+1)^4
$$
and on the left-hand side we have
$$
(k-1)^4 + 4k^3 = k^4 + 6k^2-4k+1 = k^4 + 1 + 2k(3k-2)\geq k^4
$$
(where that last inequality is true because both $1$ and $2k(3k-2)$ are positive).
Gathering it all up, this yields
$$
\frac{k^4}4\leq \frac{(k-1)^4 + 4k^3}4\leq 1+\cdots +(k-1)^3 + k^3\leq\frac{k^4+4k^3}4\leq \frac{(k+1)^4}4
$$
and this finishes the induction step.
