If x is real, and $$x^4 - 5x^2 - 1 = 0$$, then the value of $$\left(x^6 - 3x^2 + \frac{3}{x^2} - \frac{1}{x^6} + 1\right)$$ is:

$$x^4 - 5x^2 - 1 = 0$$

Multiply by $$\frac{1}{x^2}$$ in the above equation.

$$\frac{1}{x^2}(x^4 - 5x^2 - 1 = 0)$$

$$x^2-5-\frac{1}{x^2}=0$$

$$x^2-\frac{1}{x^2}=5$$

Cube of the above given equation.

$$\left(x^2-\frac{1}{x^2}\right)^3=5^3$$

$$x^6-\frac{1}{x^6}-3\times\ x^2\times\ \frac{1}{x^2}\left(x^2-\frac{1}{x^2}\right)=125$$

$$x^6-\frac{1}{x^6}-3\left(x^2-\frac{1}{x^2}\right)=125$$

$$x^6-\frac{1}{x^6}-3x^2+\frac{3}{x^2}=125$$    Eq.(i)

Put the value of the Eq.(i) in the below equation.

value of $$\left(x^6 - 3x^2 + \frac{3}{x^2} - \frac{1}{x^6} + 1\right)$$ = (125+1) = 126