<s>[INST] Prove by induction $\sum_{i=1}^n i^3 = \frac{n^2(n+1)^2}{4}$ In this problem I had to use the method of mathematical induction in order to solve $$\sum_{i=1}^n i^3 = \frac{n^2(n+1)^2}{4}$$
After proving that $P_1$ is true, and writing out $P_k$ I proceeded to solve for $P_{k+1}$. I wrote it out, using $P_k$ to substitute, as $$\frac{k^2(k+1)^2}{4}+\frac{4}{4}(k+1)^3=\frac{(k+1)^2(k+2)^2}{4}$$
I continued expanding on the LHS:=
$$\frac{k^2(k^2+2k+1)+(4k+4)^3}{4}=$$
$$\frac{k^4+2k^3+k^2+(4k+4)(16k^2+32k+16}{4}=$$
$$\frac{k^4+68k^3+192k^2+192k+64}{4}=$$
And on the RHS I did the same:
$$=\frac{(k+1)^2(k+2)^2}{4}$$
$$=\frac{(k^2+2k+1)(k^2+4k+4)}{4}$$
When I make the LHS=RHS I get equations that arent equal to each other. 
$$\frac{k^4+68k^3+192k^2+192k+64}{4}=\frac{(k^2+2k+1)(k^2+4k+4)}{4}$$
Any ideas?
EDIT: As pointed out by LordSharktheUnknown, $4(k+1)^3≠(4k+4)^3$, currently working on it to see if this will resolve the problem.
 [/INST] We have that
$$\sum_{i=1}^{n+1} i^3=\frac{n^2(n+1)^2}{4}+\frac{4}{4}(n+1)^3=\frac{(n+1)^2(n^2+4n+4)}{4}=\frac{(n+1)^2(n+2)^2}{4}$$
</s>