We have the sum of cubes identity
[tex]a^3 + b^3 = (a + b) (a^2 - ab + b^2)[/tex]
and observing that 1 + 4 = 2 + 3, we have
[tex]1^3 + 4^3 = (1 + 4) (1^2 - 4 + 4^2) = 5 \times 13[/tex]
and
[tex]2^3 + 3^3 = (2 + 3) (2^2 - 6 + 3^2) = 5 \times 7[/tex]
Then
[tex]\sqrt{1^3+2^3+3^3+4^3+5^3} = \sqrt{5\times13 + 5\times7 + 5\times5^2} = \sqrt{5 \times (20 + 25)} \\\\ ~~~~~~~~ = \sqrt{5^2 \times (4 + 5)} = \sqrt{5^2\times9} = \sqrt{5^2\times3^2} = 5\times3 = \boxed{15}[/tex]
Alternatively, we have the well-known sum of cubes formula
[tex]\displaystyle \sum_{i=1}^n i^3 = \frac{n^2(n+1)^2}4[/tex]
The sum under the square root is this sum with [tex]n=5[/tex]. Then
[tex]1^3+2^3+3^3+4^3+5^3 = \dfrac{5^2\times6^2}4 = 225 = 15^2[/tex]
and so the square root again reduces to 15.
