If (x + y + z) = 12, xy + yz + zx = 44 and xyz = 48, then what is the value of $$x^3 + y^3 + z^3$$ ?

Given : $$xy+yz+zx=44$$ and $$xyz=48$$ ---------------(i)

and $$x+y+z=12$$ -------------(ii)

Squaring both sides, we get :

=> $$(x+y+z)^2=(12)^2$$

=> $$(x^2+y^2+z^2)+2(xy+yz+zx)=144$$

=> $$(x^2+y^2+z^2)+2(44)=144$$

=> $$x^2+y^2+z^2=144-88=56$$ -------------(iii)

We know that, $$x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2-xy-yz-zx)$$

Substituting values from equations (i),(ii) and (iii),

=> $$x^3+y^3+z^3 - 3(48) = (12)(56-44)$$

=> $$x^3+y^3+z^3-144=144$$

=> $$x^3+y^3+z^3=144+144=288$$

=> Ans - (D)