Sum of cubes of first n natural numbers is given by below formula
\[\sum_{1}^{n} n^{3}=(\frac{n(n+1)}{2})^{2} \]
where n is natural number
This method can easily be employed to calculate the cubes of n natural numbers using the above formula
For example \[\sum_{1}^{5} n^{3}= 1^{3}+2^{3}+3^{3}+4^{3}+5^{3} \]
\[\sum_{1}^{5} n^{3}= 1^{3}+2^{3}+3^{3}+4^{3}+5^{3}=[\frac{5*6}{2}]^{2}=225 \]
\[\sum_{1}^{10} n^{3}= 1^{3}+2^{3}+3^{3}+4^{3}+5^{3}+......+10^{3}=[\frac{10*11}{2}]^{2}=3025 \]
\[\sum_{1}^{20} n^{3}= 1^{3}+2^{3}+3^{3}+4^{3}+5^{3}+......+20^{3}=[\frac{20*21}{2}]^{2}=44100 \]
and so on.
