$\begin{align}  &\sum\nolimits_{r=1}^{n}{{{r}^{3}}}={{\left[ \tfrac{1}{2}n(n+1) \right]}^{2}} \\ & {{1}^{3}}+{{2}^{3}}+{{3}^{3}}+\cdot \cdot \cdot +{{n}^{3}}={{\left[ \tfrac{1}{2}n(n+1) \right]}^{2}} \\ & \text{when }n=1 \\ & L.H.S={{1}^{3}} \\ & R.H.S.={{[\tfrac{1}{2}(1)(1+1)]}^{2}}=1 \\ & \text{when }n=2 \\ & L.H.S \\ & {{1}^{3}}+{{2}^{3}}=9 \\ & R.H.S=[\tfrac{1}{2}(2){{(2+1)}^{2}}]=9 \\ & \text{The preposition hold true for }n\text{ }term \\ & \text{Let }n=k \\ & {{1}^{3}}+{{2}^{3}}+{{3}^{3}}+\cdot \cdot \cdot +{{k}^{3}}={{\left[ \tfrac{1}{2}k(k+1) \right]}^{2}} \\ & \text{For }k+1\text{ terms } \\ & \underbrace{{{1}^{3}}+{{2}^{3}}+{{3}^{3}}+\cdot \cdot \cdot +{{k}^{3}}}_{{{\left[ \tfrac{1}{2}k(k+1) \right]}^{2}}}+{{(k+1)}^{3}}={{\left[ \tfrac{1}{2}k(k+1) \right]}^{2}}+{{(k+1)}^{3}} \\ & ={{(k+1)}^{2}}\left[ \frac{{{k}^{2}}}{4}+(k+1) \right] \\ & ={{(k+1)}^{2}}\left( \frac{{{k}^{2}}+4k+4}{4} \right) \\ & ={{(k+1)}^{2}}{{\left( \frac{k+2}{2} \right)}^{2}}={{\left( \frac{(k+1)(k+2)}{2} \right)}^{2}} \\ & \text{The preposition hold true for }k+1\text{ terms, therefore, the formula is valid}\text{.} \\\end{align}$