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L B H = V
[ x + 2 ] [ 3x ] [ x ] = 288
[ x + 2 ] [ 3x² ] = 288
3x³ + 6x² - 288 = 0
x = 4
L = 6 ins , W = 12 ins , H = 4 ins

The width of a rectangular box is 3 times its height, and its length is 2 more than its height.
Find the dimensions of the box if its volume is 288 cubic inches.
V
= W * L * H
= 3H * (2 + H) * H
= 6H^2 + 6H^2
= 3 H^3 + 6 H^2
3 H^3 + 6 H^2 = 288
H^3 + 2H^2 = 96
H = 4
Dimensions of the box:
Length: 6 inches
Width: 12 inches
Height: 4 inches

Let h represent the height. Then the width is 3h, and the length is h+2. The volume is
.. 3h^2(h+2) = 288
This equation has one real solution at h = 4.

The box dimensions are …
.. 4 in high, 12 in wide, 6 in long

w = 3h
l = 2 + h
v = l * w * h
v = 288

288 = 3h * (2 + h) * h
288/3 = h^2 * (h + 2)
96 = h^3 + 2h^2

Possible rational values for h (since we're dealing with real measurements, then we can restrict this to positive values)

h = 1 , 2 , 3 , 4 , 6 , 8 , 12 , 16 , 24 , 32 , 48 , 96

We need to restrict this to values where h^3 < 96, so that limits us to 1 , 2 , 3 , and 4

96 = h^3 + 2h^2
96 = 1 + 2 = 3
96 = 8 + 2 * 4 = 8 + 8 = 16
96 = 3^3 + 2 * 3^2 = 27 + 18 = 45
96 = 4^3 + 2 * 4^2 = 64 + 32 = 96

h = 4
w = 3h = 12
l = h + 2 = 6