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,
$$\Rightarrow x^3 + y^3 + z^3=(19)(133-114)+3\times 216$$
$$\Rightarrow x^3 + y^3 + z^3=(19)(133-114)+648$$
$$\Rightarrow x^3 + y^3 + z^3=(19)(19)+648$$
$$\Rightarrow x^3 + y^3 + z^3=361+648=1009$$
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