Do you have the expression x^2 + y^2 + z^2 + 36
or do you have the equation
x^2 + y^2 + z^2 = 36
The last would be a sphere with radius 6 and centre at (0,0,0) in 3-D
What is being "square rooted" ??
What are you trying to do, find ordered triples that satisfy the equation???
btw, 10, 8, and 5 do not satisfy the equation.
2= squared
x2+y2+z2+36 then square rooted and then the number must be rational.
I need x y and z. I already used 10 8 and 5
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