Abox has a volume of 192 cubic inches, a length that is twice as long as is width, and a height that is 2 inches greater than the width. what are the dimensions of the box

Volume of box has a formula of: length × width × height

We know from the question that the length is 2 times of the width, and the height is 2inches greater than the width.

Since all 3 sides are relative to the width, let x be the width,
Width = x inches
Length = 2x inches
Height = (x + 2) inches

We know that Volume = Length × Width × Height:
192 inches^3 = 2x × x × (x + 2) inches
192 = 2x × x × (x + 2)
192 = 2x^2 × (x + 2)
192 =2x^3 + 4x^2
{Divide both sides of equation by 2}
96 = x^3 + 2x^2
{Bring all values to one side of the equation by subtracting 96 from both sides}
0 = x^3 + 2x^2 - 96
{Factorise}
(x - 4) (x^2 + 6x + 24) = 0
{Divide both sides by (x^2 + 6x + 24). 0 divided by (x^2 + 6x + 24) will give 0}
(x - 4) = 0
x = 4
{x^2 + 6x + 24 has no real x-value, so leave it. Do let me know if you'd like me to prove that (x^2 + 6x + 24) has no real x-value :)

Recall:
Width = x inches
Width = 4 inches
Length = 2x inches
Length = 2 × 4 inches
= 8 inches
Height = x + 2 inches
Height = 4 + 2 inches
= 6 inches

Hope this helps! :)