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## Using the principle of mathematical induction to show identities III

Solution: The $n$-th proposition is
$$P_n: 1^3+2^3+\cdots+n^3=(1+2+\cdots+n)^2.$$ Then $P_1$ asserts $1^3=1^2$ which is clearly true and we have the induction basis.

Now we assume $P_n$ is true, that is the equation
\begin{equation}\label{eq:1-1-3-1}
1^3+2^3+\cdots+n^3=(1+2+\cdots+n)^2.
\end{equation} We would like to show $P_{n+1}$ is true based on $P_n$. To do that, we shall use the following identity from Example 1
\begin{equation}\label{eq:1-1-3-2}
1+2+\cdots+n=\frac{1}{2}n(n+1).
\end{equation} We add both sides of \eqref{eq:1-1-3-1} by $(n+1)^3$ and obtain
\begin{align*}
&\ 1^3+2^3+\cdots +n^3+(n+1)^3\\
=&\ (1+2+\cdots+n)^2+(n+1)^3\\
\text{use \eqref{eq:1-1-3-2}}=&\ \frac{1}{4}n^2(n+1)^2+(n+1)^3\\
=&\ \frac{1}{4}n^2(n+1)^2+\frac{1}{4}(n+1)^2(4n+4)\\
=&\ \frac{1}{4}(n+1)^2(n^2+4n+4)\\
=&\ \frac{1}{4}(n+1)^2(n+2)^2\\
=&\ \left(\frac{1}{2}(n+1)\big((n+1)+1\big)\right)^2\\
\text{use \eqref{eq:1-1-3-2}}=&\ \big(1+2+\cdots+n+(n+1)\big)^2.
\end{align*} Therefore $P_{n+1}$ is true if $P_n$ is true. By the principle of mathematical induction, we make the conclusion that $P_n$ is true for all positive integers $n$.