**Question:**

Find the sum $6^{3}+7^{3}+8^{3}+9^{3}+10^{3}$

**Solution:**

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 .

$6^{3}+7^{3}+8^{3}+9^{3}+10^{3}=3025-225=2800$