The sum of three consecutive whole numbers is equal to the cube of the square root of the middle number. Find the three numbers

Step 1: Represent the three consecutive whole numbers. Let the middle number be $n$. Then the three consecutive whole numbers are $n-1$, $n$, and $n+1$. Since they are whole numbers, $n-1 \ge 0$, which means $n \ge 1$.

Step 2: Set up the equation based on the given condition. The sum of the three numbers is $(n-1) + n + (n+1)$. The cube of the square root of the middle number is $(\sqrt{n})^3$. So, the equation is: $(n-1) + n + (n+1) = (\sqrt{n})^3$

Step 3: Simplify and solve the equation for $n$. Simplify the left side: $3n = (\sqrt{n})^3$ We can rewrite $(\sqrt{n})^3$ as $n\sqrt{n}$. $3n = n\sqrt{n}$ Since $n \ge 1$, we can divide both sides by $n$: $3 = \sqrt{n}$ Square both sides to solve for $n$: $3^2 = (\sqrt{n})^2$ $9 = n$

Step 4: Find the three numbers. Substitute $n=9$ back into our expressions for the three numbers: First number: $n-1 = 9-1 = 8$ Middle number: $n = 9$ Third number: $n+1 = 9+1 = 10$

Step 5: Verify the solution. The three numbers are 8, 9, 10. All are whole numbers. Sum: $8+9+10 = 27$. Middle number: $9$. Cube of the square root of the middle number: $(\sqrt{9})^3 = 3^3 = 27$. The sum (27) is equal to the cube of the square root of the middle number (27). The condition is satisfied.

The three numbers are $\boxed{8, 9, 10}$.