<s>[INST] How to prove that $\forall n\in \mathbb{N}$, $\sum ^{n}_{i=1}i^{3}=\frac {n^{2}(n+1)^{2}}{4}$? Use mathematical induction to prove that $\forall n\in \mathbb{N}$,
$$\sum ^{n}_{i=1}i^{3}=\dfrac {n^{2}(n+1)^{2}}{4}$$
$$\begin{align*}
\sum_{k=1}^{n+1} k^3 &= \sum_{k=1}^{n} k^3 + (n+1)^2 \stackrel{\rm(IH)}{=} \dfrac {n^{2}(n+1)^{2}}{4} + (n+1)^2 \\
&= \dfrac {n^{2}(n+1)^{2}+4(n+1)^2}{4}
\end{align*}$$
Is it true? What to do next?
 [/INST] First, don't forget the base case, $n = 1$: It holds.
Your inductive hypothesis is fine, but your inductive step is off.
You want: $$\begin{align*}
\sum_{k=1}^{n+1} k^3 &= \left(\sum_{k=1}^{n} k^3\right) + (n+1)^{\color{blue}{\bf 3}} \stackrel{\rm(IH)}{=} \dfrac {n^{2}(n+1)^{2}}{4} + (n+1)^\color{blue}{\bf 3} \\
&= \dfrac {n^{2}(n+1)^{2}+4(n+1)^3}{4}\\
&= \frac{(n+1)^2(n^2 + 4(n+1))}{4}\\
&=\frac{(n+1)^2 (n^2 + 4n + 4)}{4}\\
& = \frac{(n+1)^2(n+2)^2}{4}
\end{align*}$$
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