Find the sum $6^{3}+7^{3}+8^{3}+9^{3}+10^{3}$
It is required to find the sum $6^{3}+7^{3}+8^{3}+9^{3}+10^{3}$.
$6^{3}+7^{3}+8^{3}+9^{3}+10^{3}=$ Sum of cubes of natural numbers starting from 1 to 10 - Sum of cubes of natural numbers starting from 1 to 5 .
Note:
Sum of cubes of first $n$ natural numbers, $1^{3}+2^{3}+3^{3}+\ldots \ldots n^{3}$,
$\sum_{k=1}^{n} k^{3}=\left(\frac{n(n+1)}{2}\right)^{2}$
From the above identities,
Sum of cubes of natural numbers starting from 1 to 10
$=\left(\frac{10(11)}{2}\right)^{2}=3025$
Sum of cubes of natural numbers starting from 1 to 5
$=\left(\frac{5(6)}{2}\right)^{2}=225$
$6^{3}+7^{3}+8^{3}+9^{3}+10^{3}=$ Sum of cubes of natural numbers starting from 1 to $10-$ Sum of cubes of natural numbers starting from 1 to 5 .
$6^{3}+7^{3}+8^{3}+9^{3}+10^{3}=3025-225=2800$