# precalculus algebra – Factoring in an induction test I have to try the following:

$$1 + 3 ^ 3 + … + (2n + 1) ^ 3 = (n + 1) ^ 2 (2n ^ 2 + 4n + 1)$$ Inductively.

My intent:

Base case, $$n = 1$$:

$$1 + 3 ^ 3 = (2) ^ 2 (2 cdot1 ^ 2 + 4 cdot1 + 1)$$, which is true.

By inductive hypothesis, suppose $$n = k$$:

$$1 + 3 ^ 3 + … + (2k + 1) ^ 3 = (k + 1) ^ 2 (2k ^ 2 + 4k + 1)$$

by $$n = k + 1$$

$$1 + 3 ^ 3 + … + (2k + 1) ^ 3 + (2k + 3) ^ 3 = (k + 2) ^ 2 (2 (k + 1) ^ 2 + 4 (k + 1) + 1)$$

Using the inductive hypothesis we need to prove that

$$(k + 1) ^ 2 (2k ^ 2 + 4k + 1) + (2k + 3) ^ 3 = (k + 2) ^ 2 (2 (k + 1) ^ 2 + 4 (k + 1) + 1)$$

And here I have the problem, because I do not know how to manipulate any of the sides of the equation to prove this. I tried it on both sides but I can not find the way. One of my last attempts ended here:

$$(k + 1) ^ 2 (2k ^ 2 + 4k + 1) + (2k + 3) ^ 3 = (k + 2) ^ 2 (2 (k + 1) (k + 3) + 1)$$

Which is true for alpha wolfram.

PS: I am aware that there are many documents and information about this test, but I am not looking for another form or something, because I saw many publications on how to prove this claim through induction, but all were used. the last term is (2n-1) ^ 3, and I need to prove it when it is (2n + 1) ^ 3, and the final expression is a bit different. I just need help factoring my last step.

Corrections in the inductive steps that I followed are also appreciated. 