If x + y + z = 19, xy + yz + zx = 114, then the value of $$\sqrt{x^3 + y^3 + z^3 - 3xyz}$$ is:

Given that,
x + y + z = 19, xy + yz + zx = 114

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

Hence, $$(x+y+z)^2-2(xy+yz+zx)=x^2+y^2+z^2$$

Now, substituting the values,  $$(19)^2-2(114)=x^2+y^2+z^2$$

$$\Rightarrow 361-228=x^2+y^2+z^2$$

$$\Rightarrow x^2+y^2+z^2=133$$

Now, $$\Rightarrow \sqrt{x^3 + y^3 + z^3 - 3xyz}=\sqrt{(x+y+z)(x^2+y^2+z^2-xy-yz-zx)}$$

Now, substituting the values in the equation,

$$\Rightarrow \sqrt{x^3 + y^3 + z^3 - 3xyz}=\sqrt{(19)(133-114)}=\sqrt{19\times 14}=19$$