Question

Can you see an easy way to work out the value of
`\sqrt{ 1^3 + 2^3 + 3^3 + 4^3 + 5^3` ? if so, describe it.

Answer 1

We have the sum of cubes identity

`a^3 + b^3 = (a + b) (a^2 - ab + b^2)`

and observing that 1 + 4 = 2 + 3, we have

`1^3 + 4^3 = (1 + 4) (1^2 - 4 + 4^2) = 5 * 13`

and

`2^3 + 3^3 = (2 + 3) (2^2 - 6 + 3^2) = 5 * 7`

Then

`√(1^3+2^3+3^3+4^3+5^3) = √(5*13 + 5*7 + 5*5^2) = √(5 * (20 + 25)) \\\\ ~~~~~~~~ = √(5^2 * (4 + 5)) = √(5^2*9) = √(5^2*3^2) = 5*3 = \boxed{15}`

Alternatively, we have the well-known sum of cubes formula

`\displaystyle \sum_(i=1)^n i^3 = \frac{n^2(n+1)^2}4`

The sum under the square root is this sum with
`n=5`. Then

`1^3+2^3+3^3+4^3+5^3 = \frac{5^2*6^2}4 = 225 = 15^2`

and so the square root again reduces to 15.