There must be a better way to solve this but this is what I did:

If we call the side length x, and the coordinates of the point in space relative to the middle vertex, (a,b,c) we can form four equations:

a^2 + b^2 + c^2 = 583^2

(a+x)^2 + b^2 + c^2 = 637^2

a^2 + (b+x)^2 + c^2 = 713^2

a^2 + b^2 + (c+x)^2 = 727^2

Subtracting the first equation from each of the other three in turn gives:

2ax + x^2 = 65880

2bx + x^2 = 168480

2cx + x^2 = 188640

Rearrange each to isolate a,b,c

a = (65880 - x^2)/2x = 32940/x - x/2

b = (168480 - x^2)/2x = 84240/x - x/2

c = (188640 - x^2)/2x = 94320/x - x/2

Let’s square each of those expressions

a^2 = 32940^2/x^2 – 32940 + x^2/4

b^2 = 84240^2/x^2 – 84240 + x^2/4

c^2 = 94320^2/x^2 – 94320 + x^2/4

Adding them together and equating the sum to 583^2:

(32940^2+84240^2+94320^2)/x^2 – 211500 + 3(x^2)/4 = 583^2

Multiply every term by 4x^2 to eliminate the x^2 term in the denominator:

Solve the quadratic in x^2 to x^2 = 32400 or 2108356/3, so x is either 180 or 838.322929…, however in this second case the point would be inside the cube.

The cube’s side length is therefore 180