Answer

Prove: $1^3 + 2^3 + \dots + n^3 = \left(\dfrac{n(n+1)}{2}\right)^2$ for all $n \in \mathbb{N}$ by Mathematical Induction.

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

LHS $= 1^3 = 1$

RHS $= \left(\dfrac{1\cdot 2}{2}\right)^2 = 1^2 = 1$

LHS = RHS, so holds for $n=1$.

Step 2: Induction Hypothesis

Assume true for $n=k$:
$$P(k): 1^3 + 2^3 + \dots + k^3 = \left(\frac{k(k+1)}{2}\right)^2$$

Step 3: Inductive Step ($n = k+1$)

To prove $P(k+1)$:
$$1^3 + \dots + k^3 + (k+1)^3 = \left(\frac{(k+1)(k+2)}{2}\right)^2$$

Consider LHS of $P(k+1)$:
$$\begin{aligned} \text{LHS} &= \left(1^3 + \dots + k^3\right) + (k+1)^3 \\ &= \left(\frac{k(k+1)}{2}\right)^2 + (k+1)^3 \quad \text{(by hypothesis)}\\ &= \frac{k^2(k+1)^2}{4} + (k+1)^3 \\ &= (k+1)^2 \left[ \frac{k^2}{4} + (k+1) \right] \\ &= (k+1)^2 \left[ \frac{k^2 + 4k +4}{4} \right] \\ &= (k+1)^2 \cdot \frac{(k+2)^2}{4} \\ &= \left( \frac{(k+1)(k+2)}{2} \right)^2 = \text{RHS of } P(k+1) \end{aligned}$$

Thus $P(k) \implies P(k+1)$.

Step 4: Conclusion

By the Principle of Mathematical Induction, since base case holds and inductive step proved, the formula holds for all $n \in \mathbb{N}$.

Corollary / Remark

Note that $1+2+\dots+n = n(n+1)/2$, so the theorem elegantly states:
$$1^3 + 2^3 + \dots + n^3 = (1+2+\dots+n)^2$$