Mathematical Induction: Addition of uneven cubes

Question

The following mathematical induction problem about uneven cube roots seems to be a common challenge amongst college students:

$$1^3+3^3+5^3+...+(2n-1)^3=n^2(2n^2-1)$$

Other sources, like here, provide, explain and elaborate on the sum of consecutive cube roots:

$$1^3+2^3+3^3+...+n^3=\frac{n^2(n^2+1)^2}{4}$$

The purpose of both problems is to prove the formulae using mathematical induction.
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Here are the first few steps used to prove the first formula:

$$1^3+3^3+5^3+...+(2n-1)^3=n^2(2n^2-1)$$

Step 1 Prove that both sides are equal by replacing the constant 1 into the equation:

$$(2n-1)^3=n^2(2n^2-1)$$ $$(2(1)-1)^3=(1)^2(2(1)^2-1)$$ $$1=1$$

Step 2 Now that we know that both sides are equal to one, assume the following (replace n by k):

$$1^3+3^3+5^3+...+(2k-1)^3=k^2(2k^2-1)$$

Step 3 Using the beforementioned; prove the formula as follows:

$$1^3+3^3+5^3+...+(2k-1)^3+(2(k+1)-1)^3=(k+1)^2(2(k+1)^2-1)$$

We know from step 2 that:

$$1^3+3^3+5^3+...+(2k-1)^3=k^2(2k^2-1)$$

Therefore:

$$k^2(2k^2-1)+(2(k+1)-1)^3=(k+1)^2(2(k+1)^2-1)$$ $$k^2(2k^2-1)+(2k+1)^3=(k+1)^2(2(k+1)^2-1)$$

Step 4 Manipulate the formula in order to prove that both sides of the equation are equal (the same). Thus, manipulating the left side to get the right side will prove the formula using mathematical induction:

$$k^2(2k^2-1)+(2k+1)^3$$ $$= ...$$ $$=...$$ $$=(k+1)^2(2(k+1)^2-1)$$

The complication arises here (above). Please assist on how the left side can be manipulated into finding the exact copy of the right side.

Answer 1

Starting from where you got to:

$$k^2(2k^2-1)+(2(k+1)-1)^3$$ $$=2k^4-k^2+(2k+1)^3$$ $$=2k^4-k^2+(8k^3+12k^2+6k+1)$$ $$=2k^4+8k^3+11k^2+6k+1$$ $$=(k+1)(2k^3+6k^2+5k+1)$$ $$=(k+1)(k+1)(2k^2+4k+1)$$ $$=(k+1)^2(2((k+1)-1)^2+4((k+1)-1)+1)$$ $$=(k+1)^2(2((k+1)^2-2(k+1)+1)+4(k+1)-3)$$ $$=(k+1)^2(2(k+1)^2-4(k+1)+2+4(k+1)-3)$$ $$=(k+1)^2(2(k+1)^2-1)$$

As required

feel free to ask if you need further clarification

EDIT: You asked how I did the factorization. Firstly, I knew that we had to reach the form $(k+1)^2(2(k+1)^2-1)$, so I knew I needed to pull out a factor of $(k+1)$. From that assumption, I essentially reverse-engineered the factorization:

$$2k^4+8k^3+11k^2+6k+1=(k+1)(...)$$

As our product is 4th order (contains $k^4$) and $(k+1)$ is first order, the other part of the factorization must be 3rd order (ie: highest power of $k$ is $k^3$). We also know that the $k^3$ term and the $k$ term in $(k+1)$ has to multiply to $2k^4$ (as there is no other way to reach a $k^4$ as all other terms have smaller powers of $k$). Ultimately, that means that the first term is $2k^3$, as $2k^3*k=$2k^4$

$$2k^4+8k^3+11k^2+6k+1=(k+1)(2k^3+...)$$

Now, we know that when the $2k^3$ term multiplies the $1$ term from $(k+1)$, we'll get $2k^3$. We also know that the only other way to get a $k^3$ term is to multiply the $k^2$ term by the $k$ term from $k+1$. As we know the $k^3$ terms eventually end up as $8k^3$, we can see that the $k^2$ term in the factorization must be $6k^2$, as when we expand, we'll end up with a $k^3$ term of $6k^3+2k^3=8k^3$.

$$2k^4+8k^3+11k^2+6k+1=(k+1)(2k^3+6k^2+...)$$

We use the same logic for the $k$ term. We need to reach $11k^2$, and we already have $6k^2$ from our $k^2$ term, so our $k$ term must be $5k$. Similarly for the unit term. We need to reach $6k$, and our $k$ term gives us $5k$, so our unit term is just $1$

$$2k^4+8k^3+11k^2+6k+1=(k+1)(2k^3+6k^2+5k+1)$$

We can see that we've factorized correctly as the unit term has come out correctly without us having to calculate it. If the unit term hadn't matched, it would either mean that we made a mistake, or that our polynomial wasn't divisible by $k+1$.