Cube In A Cube

let $x$ be the side length of the cube-shaped container

The volume of rise of water is equal to the volume of the cube that was dropped.

So, we have

$252x^2=1008^3$

It follows that,

$x^2=\dfrac{1008^3}{252}=4064256$

Finally,

$x=\sqrt{4064256}=$ $\color{#D61F06}\boxed{\large 2016}$

Let the side length of the cube-shaped container be $x$ . Then, we have this equation $252x^2=1008^3$ Solving, we get $x=2016$ .

Volume = Base Area * Height; Since the displaced volume of the fluid = the volume of the submerged cube; and since 1008 (the submerged cube's height) = 4 * 252 (the displacement of the fluid level); therefore, the area of the bigger cube must be equal to 4 times the area of the smaller cube; which means the side length of the bigger cube = 2 times the side length of the smaller one = 2*1008 = 2016.

Let S be the length of the side of the cube shaped container.

Let D be the depth that it is filled before the smaller cube is put in.

Let C = $(1008^3)$ , the volume of the smaller cube.

Initial filled volume is $S^2 \times D$

After the smaller cube is submerged the filled volume is $S^2 \times (D + 252) = (S^2 \times D) + C$

Expand $S^2 \times (D + 252)$ to get

$(S^2 \times D) + ( S^2 \times 252) = (S^2 \times D) + C$

Subtract $(S^2 \times D)$ from both sides to get

$S^2 \times 252 = C$ and

$S = \sqrt { C / 252 } = \sqrt { 1,024,192,512 / 252 } = 2016$

In addition, we can see that this problem has probably been around the site for a while now.

The volume of water that rises in the cube-shaped container is equal to the volume of the cube with edge length of 1008 units. Therefore,

$\color{#3D99F6}1008^3=\text{area of the base} \times \text{height} = \text{area of the base} \times 252$

$\color{plum}\implies$ $\color{#3D99F6}\text{area of the base}=4064256$

$\color{plum}\implies$ $\color{#3D99F6}\text{length of a side of the cube-shaped container} = \sqrt{\text{area of the base}}=\sqrt{4064256} = \color{#D61F06}\boxed{2016}$