ethanyessir

Well there's two cubes, so the smallest possible configuration of them would be a rectangular prism...

So we have a rectangular prism with height 1, width 1, and length 2. 

Surface area would be 10.

Correct me if im wrong.

Use 3d pythagorean theorem,

3^2+5^2+11^2=x^2.

9+25+121=x^2

155=x^2

x=sqrt 155

So the distance between the two opposite corners is sqrt 155

OK so,

lets first rearrange the second equation: x^2-xy+y^2=64.

and we know that x+y=10.

we can rewrite the second equation like this

(x+y)^2-3xy=64

and since x+y=10,

100-3xy=64

which means that xy=12.

NOW

we solve x+y=10 and xy=12.

we can substitute x as 10-y

so we have

y(10-y)=12

10y-y^2=12

and we get the quadratic

y^2-10y+12=0.

this isnt easily factorable so i will use the quadratic formula.

we get that y=5+sqrt 13, 5-sqrt 13.

so that means 

(x,y)=(5+sqrt 13, 5-sqrt 13) and (5-sqrt 13, 5+sqrt 13).

:)

Hey,

I got the equation: x^3-3x^2+12x+10=0.

The problem is that this equation is not factorable with rational numbers and I don't think they want you to use the cubic formula.

Is there something wrong with the problem?