If n is an odd integer greater than or equal to 1 , the value of  $n^3-{(n-1)}^3+{(n-2)}^3-\dots +{(-1)}^{n-1}{1}^3$  is

Given that n is an odd integer greater than or equal to 1 .
$$\begin{gathered} S_n=n^3-{(n-1)}^3+{(n-2)}^3-\dots +{(-1)}^{n-1}{1}^3 \\ =1^3-2^3+\dots +{(n-2)}^3-{(n-1)}^3+n^3 \end{gathered}$$
  
For[ $\because n$  is odd integer, so  $(n-1)$ is even integer] $=(1^3+2^3+\dots +n^3)-2·2^3(1^3+2^3+\dots +\frac{n-1}{2}\text{ terms })$So,
$$\begin{gathered} ={\left[\frac{n(n+1)}{2}\right]}^3-16·{\left[\frac{\frac{n-1}{2}(\frac{n-1}{2}+1)}{2}\right]}^3 \\ =\frac{n^2{(n+1)}^2}{4}-\frac{4{(n-1)}^2{(n+1)}^3}{16}=\frac{{(n+1)}^2}{4}[n^2-{(n-1)}^2] \\ =\frac{{(n-1)}^3}{4}+(2n-1)\left(1\right)=\frac{(2n-1){(n+1)}^2}{4} \end{gathered}$$