From the question,
The formula for the volume(V1) of the cube is,
[tex]V_1=l^3[/tex]Given that:
The length of the cube is x
Therefore,
[tex]V_1=x^3[/tex]The formula for the volume (V2) of the cylinder is,
[tex]V_2=\pi r^2h[/tex]Given:
[tex]\begin{gathered} r=\frac{\text{diameter}}{2}=\frac{x}{2} \\ h=\text{height}=x \end{gathered}[/tex]Therefore,
[tex]\begin{gathered} V_2=\pi\times(\frac{x}{2})^2\times x=\pi\times\frac{x^2}{4}\times x=\frac{\pi x^3}{4} \\ \therefore V_2=\frac{\pi x^3}{4} \end{gathered}[/tex]Hence, the volume(V) of water that can be poured inside the rectangular prism yet outside the cylinder will be
[tex]\begin{gathered} V=x^3-\frac{\pi x^3}{4} \\ \therefore V=x^3-\frac{\pi}{4}x^3 \end{gathered}[/tex]Therefore, the expression that will help Cole solve for the volume is,
[tex]x^3-\frac{\pi}{4}x^3[/tex]