If $$x^2 + y^2 + z^2 = 133, xy + yz + zx = 114$$ and xyz = 216, then the value of $$x^3 + y^3 + z^3$$ is:

Given that,

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

$$xy + yz + zx = 114$$

$$xyz = 216$$

We know that $$(a+b+c)^2=a^2+b^2+c^2+2(ab+bc+ca)$$

$$\Rightarrow (x+y+z)=\sqrt{x^2+y^2+z^2+2(xy+yz+zx)}$$

$$\Rightarrow (x+y+z)=\sqrt{133+2\times114}$$

$$\Rightarrow (x+y+z)=\sqrt{361}=19$$

Hence,

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

Now, substituting the values,

$$\Rightarrow x^3 + y^3 + z^3=(19)(133-114)+3\times 216$$

$$\Rightarrow x^3 + y^3 + z^3=(19)(133-114)+648$$

$$\Rightarrow x^3 + y^3 + z^3=(19)(19)+648$$

$$\Rightarrow x^3 + y^3 + z^3=361+648=1009$$