Prove by mathematical induction that $1^3 + 2^3 + 3^3$ ... + $n^3 = [n(n+1)/2]^2$.

Step 1: Base Case ($n=1$)

We begin by verifying the base case for $n=1$. We substitute $n=1$ into both sides of the given equation and check if they are equal. The left side becomes $1^3 = 1$. The right side becomes $\left[\frac{1(1+1)}{2}\right]^2 = \left[\frac{1(2)}{2}\right]^2 = \left[1\right]^2 = 1$. Since both sides are equal, the base case holds true.

Step 2: Inductive Hypothesis

Next, we assume that the statement P(k) is true for some arbitrary positive integer k. This means we assume that the sum of the first k cubes is equal to $\left[\frac{k(k+1)}{2}\right]^2$. This assumption is called the inductive hypothesis.

Step 3: Inductive Step (P(k+1))

Now, we need to prove that the statement P(k+1) is true, assuming P(k) is true. This means we need to show that the sum of the first k+1 cubes is equal to $\left[\frac{(k+1)(k+2)}{2}\right]^2$.

Step 4: Substitute Inductive Hypothesis

We start with the left-hand side of the P(k+1) statement. Using our inductive hypothesis, we replace the sum $1^3 + 2^3 + \dots + k^3$ with its assumed equivalent, $\left[\frac{k(k+1)}{2}\right]^2$. Then we add the $(k+1)^{th}$ term, which is $(k+1)^3$.

Step 5: Factor and Simplify

We factor out $(k+1)^2$ from the expression. Then, we find a common denominator for the terms inside the square brackets and combine them. The numerator $k^2 + 4k + 4$ is a perfect square trinomial, which can be factored as $(k+2)^2$. Finally, we rewrite the entire expression as a single squared term.

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