Question 1039987
It can be verified that 

{{{(x+y+z)^3 = -2(x^3+y^3+z^3) + 6xyz + 3(x+y+z)(x^2+y^2+z^2)}}} <---Equation A.

Now {{{(x+y+z)^2 = (x^2 +y^2+z^2)+2(xy + yz + xz)}}} ==> {{{12^2 = x^2 +y^2+z^2 +2*44}}}  ==> {{{x^2 +y^2+z^2 = 56}}}

Hence, substituting into Equation A...

{{{12^3 = -2*288 +6xyz +3*12*56}}}

==> 6xyz = 288, or xyz = 48.

But 48 = 2*4*6, so let x = 2, y = 4, and z = 6.

It is easily seen that this combination of values satisfies the three original equations, and that no other combination of the factors of 
48 would yield the same outcome.  Thus  x = 2, y = 4, and z = 6.