To find the volume $V$ of the box as a function of $x$, let's follow the steps described:

1. Start with the original dimensions of the cardboard:

   • Length = 10 inches
   • Width = 6 inches

2. Cut out equal squares of side $x$ from each corner:

   • The size of each square is $x \times x$.

3. Modify the dimensions of the cardboard after cutting:

   • The new length of the box after folding will be $10 - 2x$ (since squares are cut from both ends of the 10-inch side).
   • The new width of the box will be $6 - 2x$ (since squares are cut from both ends of the 6-inch side).

4. Height of the box:

   • The height of the box is equal to the side of the square cut out, which is $x$.

5. Volume $V$ of the box:

   • Volume is calculated by multiplying the length, width, and height of the box: $V(x) = (10 - 2x) \cdot (6 - 2x) \cdot x$

Now, let's expand this expression:

$V(x) = (10 - 2x)(6 - 2x)x$

Expanding the terms inside the parentheses first:

$(10 - 2x)(6 - 2x) = 60 - 20x - 12x + 4x^2 = 60 - 32x + 4x^2$

Then, multiply by $x$:

$V(x) = x(60 - 32x + 4x^2)$

Simplify:

$V(x) = 4x^3 - 32x^2 + 60x$

So, the volume $V$ of the box as a function of $x$ is:

$\boxed{V(x) = 4x^3 - 32x^2 + 60x}$

Do you want more details or have any questions?

Here are some related questions:

1. What is the maximum possible value of $x$ for this box to exist?
2. How can we find the value of $x$ that maximizes the volume of the box?
3. What would the volume be if $x = 1$?
4. How does changing $x$ affect the box's volume?
5. What are the dimensions of the box if $x = 2$?

Tip: To maximize or minimize functions, consider taking the derivative and finding the critical points!