​Simplify: $$(1^3+2^3+3^3+...+8^3)^\frac{-5}{2}$$​​

Correct option is B

Given:

​$$\left(1^3 + 2^3 + 3^3 + \dots + 8^3\right)^{-\frac{5}{2}}$$​

Formula Used:
Use the formula for the sum of cubes of the first n natural numbers:

​$$1^3 + 2^3 + 3^3 + \dots + n^3 = \left( \frac{n(n+1)}{2} \right)^2$$​

Solution:
$$\left( \sum_{k=1}^{8} k^3 \right) = \left( \frac{8 \times 9}{2} \right)^2 = (36)^2 = 1296$$​​

So expression becomes:

​$$1296^{-\frac{5}{2}} = \left(36^2\right)^{-\frac{5}{2}} = 36^{-5}$$

= (6$$^2)^{-5} = 6 ^{-10}$$​​