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,
$$\Rightarrow (4x^3 - y^3 + x^2y^2)=4\times (-3)^3-(-3)^3+(-3)^2\times(-3)^2$$
$$\Rightarrow -108+27+9\times9$$
$$\Rightarrow -108+27+81=0$$
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