## Precalculus (6th Edition) Blitzer

The value of ${{S}_{1}}=0$, ${{S}_{2}}=6$ and ${{S}_{3}}=24$, and also the statement is true.
Let us consider the statement: ${{S}_{n}}:3\text{ is a factor of }{{n}^{3}}-n$ For ${{S}_{1}}$ one has ${{1}^{3}}-1=0$ So, the above statement is true for $n=1$ as 3 is a factor of 0. For ${{S}_{2}}$ one has \begin{align} & {{2}^{3}}-2=8-2 \\ & =6 \end{align} Therefore, the above statement is true for $n=2$ as 3 is a factor of 6. For ${{S}_{3}}$ one has \begin{align} & {{3}^{3}}-3=27-3 \\ & =24 \end{align} Thus, the above statement is true for $n=3$ as 3 is a factor of 24. The values are ${{S}_{1}}=0,{{S}_{2}}=6$, and ${{S}_{3}}=24$. Also, the statement holds true.