If $$2x^2 + y^2 + 6x - 2xy + 9 = 0$$, then the value of $$(4x^3 - y^3 + x^2y^2)$$ is:

Given that,

$$2x^2 + y^2 + 6x - 2xy + 9 = 0$$

$$\Rightarrow x^2+y^2-2xy+x^2+6x+9=0$$

$$\Rightarrow (x-y)^2+(x^2+6x+9)=0$$

$$\Rightarrow (x-y)^2+(x^2+6x+3^2)=0$$

$$\Rightarrow (x-y)^2+(x+3)^2=0$$

It will be zero, if and only if the individual terms is zero.

So, $$x-y=0$$

Hence $$x=y$$

And, $$x+3=0$$,

So, $$x=-3$$

Now, substituting the values,

$$\Rightarrow (4x^3 - y^3 + x^2y^2)=4\times (-3)^3-(-3)^3+(-3)^2\times(-3)^2$$

$$\Rightarrow -108+27+9\times9$$

$$\Rightarrow -108+27+81=0$$