If x + $$\frac 1x$$​= –1, then compute: $$x^4 + \frac{1}{x^4} + 2x^2 + \frac{2}{x^2}$$​​

Correct option is A

Given:
​$$x + \frac{1}{x} = -1$$​​
Formula Used:
​$$\left(x + \frac{1}{x}\right)^2 = x^2 + \frac{1}{x^2} + 2 \\\left(x^2 + \frac{1}{x^2}\right)^2 = x^4 + \frac{1}{x^4} + 2$$​​
Solution:

$$x + \frac{1}{x} = -1\\[4pt]\text{Squaring both sides:} \\\left(x + \frac{1}{x}\right)^2 = (-1)^2 \\x^2 + \frac{1}{x^2} + 2 = 1 \\x^2 + \frac{1}{x^2} = -1\\[6pt]\text{Squaring again:} \\\left(x^2 + \frac{1}{x^2}\right)^2 = (-1)^2 \\x^4 + \frac{1}{x^4} + 2 = 1 \\x^4 + \frac{1}{x^4} = -1\\[6pt]\text{Now evaluate:} \\x^4 + \frac{1}{x^4} + 2x^2 + \frac{2}{x^2} \\= (-1) + 2(-1) \\= -3$$​