The sum of three consecutive whole numbers is equal to the cube of the square root of the middle number. Find the three 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 $n \ge 1$.

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:

Simplify the left side:

We can rewrite $(\sqrt{n})^3$ as $n\sqrt{n}$.

Since $n \ge 1$, we can divide both sides by $n$:

Square both sides to solve for $n$:

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

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 $\text{8, 9, 10}$.