If x + y = 1 and xy(xy - 2) = 12, then the value of $$x^4 + y^4$$ is:

Given that,

$$x + y = 1$$

$$\Rightarrow xy(xy-2)=12$$

$$\Rightarrow x^2 y^2-2xy=12$$

$$\Rightarrow x + y = 1$$

Now, squaring both side,

$$\Rightarrow (x + y)^2 = 1$$

$$\Rightarrow x^2+y^2+2xy = 1$$

$$\Rightarrow x^2+y^2 = 1-2xy$$

Again squaring both side,

$$\Rightarrow (x^2+y^2)^2 = (1-2xy)^2$$

$$\Rightarrow x^4+y^4+2x^2y^2 = 1+4x^2y^2-4xy$$

$$\Rightarrow x^4+y^4 = 1+4x^2y^2-4xy-2x^2y^2$$

$$\Rightarrow x^4+y^4 = 1+2x^2y^2-4xy$$

$$\Rightarrow x^4+y^4 = 1+2(x^2y^2-2xy)=1+2\times 12=25$$