The series is $$P_k: 1^3 + 3^3 + 5^3 + ... + (2k-1)^3 = k^2(2k^2-1)$$ and I have to replace $P_k$ with $P_{k+1}$ to prove the series.
I have to show that $$k^2(2k^2-1) + (2k-1)^3 = (k+1)^2[2(k+1)^2-1]$$ I'm sorry that I'm asking but their are just so many factors the algebra just passes over my head. Any help is appreciated.
Note that what you need to show is not $$k^2(2k^2-1) + (2k-1)^3 = (k+1)^2[2(k+1)^2-1].$$ What you need to show is $$k^2(2k^2-1) + (2(\color{red}{k+1})-1)^3 = (k+1)^2[2(k+1)^2-1].$$
And we can show this as the following : $$\begin{align}k^2(2k^2-1) + (2(k+1)-1)^3&=2k^4-k^2+(2k+1)^3\\&=2k^4-k^2+8k^3+12k^2+6k+1\\&=(2k^4+4k^3+2k^2)+(4k^3+8k^2+4k)+(k^2+2k+1)\\&=2k^2(k+1)^2+4k(k+1)^2+(k+1)^2\\&=(k+1)^2(2k^2+4k+1)\\&=(k+1)^2(2(k+1)^2-1).\end{align}$$