Hello Vu,
We are given to prove by induction:
$$\sum_{k=1}^n\left(k^3\right)=\left(\frac{n(n+1)}{2}\right)^2$$
First, we must check to see if our base case $P_1$ is true:
$$\sum_{k=1}^1\left(k^3\right)=\left(\frac{1(1+1)}{2}\right)^2$$
$$1^3=1^2$$
$$1=1$$
The base case is true, so we next state the induction hypothesis $P_n$:
$$\sum_{k=1}^n\left(k^3\right)=\left(\frac{n(n+1)}{2}\right)^2$$
As our inductive step, we may add $(n+1)^3$ to both sides:
$$\sum_{k=1}^n\left(k^3\right)+(n+1)^3=\left(\frac{n(n+1)}{2}\right)^2+(n+1)^3$$
On the left, incorporate the new term within the sum and factor on the right:
$$\sum_{k=1}^{n+1}\left(k^3\right)=(n+1)^2\left(\left(\frac{n}{2}\right)^2+(n+1)\right)$$
$$\sum_{k=1}^{n+1}\left(k^3\right)=(n+1)^2\left(\frac{n^2+4n+4}{4}\right)$$
$$\sum_{k=1}^{n+1}\left(k^3\right)=\frac{(n+1)^2(n+2)^2}{4}$$
$$\sum_{k=1}^{n+1}\left(k^3\right)=\left(\frac{(n+1)(n+2)}{2}\right)^2$$
$$\sum_{k=1}^{n+1}\left(k^3\right)=\left(\frac{(n+1)((n+1)+1)}{2}\right)^2$$
We have derived $P_{n+1}$ from $P_n$ thereby completing the proof by induction.