This question I answered earlier. I will like to solve in an easier method which u can follow in such problems with cubic equations. It is by successive iteration.

The INSIDE dimensions of the rectangular safe:

Length L = 1.20 m Breadth B = 1.20 m Height H = 2.00 m

Inside volume = 1.2² * 2 = 2.88 m³

Let thickness of safe be t meters

To get the outside (external) dimensions: add t on each side of a dimension (left side and right side for width and length. and at top and bottom for height). So each dimension becomes plus 2t.

Outside Length = 1.20+2t External width = 1.20+2t external height = 2.00+2t

External (Outside) volume = (1.2+2t)² (2+2t) = 2²(0.6+t)² 2 (1+t)

The volume of Steel used is the difference between the outside volume and the inside volume.

Hence : 8 (t+0.6)² (t+1) - 2.88 = 1.25 meter³

(t+0.6)² (t+1) = 4.13 / 8 = 0.51625

Let (t + 0.6) = x

x² (x+0.4) = 0.51625 or x³ + 0.4 x² - 0.51625 = 0

We can try the ways of factorization and finding zeros. But here I show another way of successive iterations to solve. rewrite the equation as,

x² = 0.51625 / (x+0.4) or x = √[0.51625 / (x+0.4)]

x = 0.7185 / √(x+0.4)

We will find the value of x by starting with some value and calculating it many times, so that we get the precision we want. Substitute a value of x on R H S and find the L H S

Start with x = 1 , new value of x = 0.7185 / √(1+0.4) = 0.6072

use x = 0.6072, new value of x = 0.7185 / √(0.6072+0.4) = 0.7159

next value of x in iteration = 0.7185 / √(0.7159+0.4) =0.68016

Next iteration x = 0.6913

next iteration x = 0.68778

next iteration x = 0.6889

next iteration x = 0.68854

next iteration x = 0.68866

next iteration x = 0.6886...

Let us take x = 0.6886 as these digits are repeated in the next iteration.

t = x - 0.6 = 0.0886 meters

or t = 8.86 cm approximately

__Verification of answer__ :

Outside volume = (1.2+2*0.0886)² (2+2*0.0886) = 4.1294 m³

Inside volume = 1.2² * 2 = 2.88 m³

Volume of steel = 1.2494 m³ ≈ 1.25 m³

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as x³ + 0.4 x² - 0.51625 = 0

x ³ = (0.51625 - 0.4 x²)

x = ∛(0.51625 - 0.4 x²)

we can use this equation also for successive iterations.