We know the following statements regarding the box:
1. The box is rectangular
2. It's twice as long as it is wide
3. The height of the box is 3 feet less than the width
4. x: feet wide
So we can apply a mathematical language to solve this problem. The volume of a rectangular box can be find by:
[tex]V=W\times L\times H \\ \\ where \\ \\ W:Width \\ L:Length \\ H:Height [/tex]
But we also know that:
[tex]W=x[/tex]
And the box is twice as long as it is wide:
[tex]L=2x[/tex]
The height of the box is 3 feet less than the width is given by:
[tex]H=x-3[/tex]
So combining these results:
[tex]V=x(2x)(x-3) \\ \therefore V=2x^2(x-3) \\ \therefore \boxed{V=2x^3-6x^2}[/tex]
The right answer is B