The problem asks us to find the value of $x$ given the dimensions of a box and its volume. The dimensions of the box are $6$, $x-4$, and $x-4$, and the volume of the box is $486$ cubic feet.

1. Problem Description

The problem asks us to find the value of $x$ given the dimensions of a box and its volume. The dimensions of the box are $6$, $x-4$, and $x-4$, and the volume of the box is $486$ cubic feet.

2. Solution Steps

The volume of a rectangular box is given by the product of its length, width, and height. In this case, the volume $V$ is given by:

$V = 6(x-4)(x-4)$

We are given that $V = 486$. Thus, we have:

$6(x-4)(x-4) = 486$

Divide both sides by 6:

$(x-4)(x-4) = \frac{486}{6}$

$(x-4)^2 = 81$

Take the square root of both sides:

$x-4 = \pm \sqrt{81}$

$x-4 = \pm 9$

We have two possible cases:

Case 1: $x-4 = 9$

$x = 9+4$

$x = 13$

Case 2: $x-4 = -9$

$x = -9+4$

$x = -5$

Since the side length of the box is $x-4$, we need $x-4 > 0$, which means $x > 4$.

If $x=13$, then $x-4 = 13-4 = 9 > 0$.

If $x=-5$, then $x-4 = -5-4 = -9 < 0$.

Since the side length cannot be negative, we must have $x=13$.

3. Final Answer

$x = 13$