If $\left(x^3 + \frac{1}{x^3} - k\right)^2 + \left(x + \frac{1}{x} - p\right)^2 = 0$ where k and p are real numbers and x ≠ 0, then $\frac{k}{p}$ is equal to:

$P^2-3$

If $\left(x^3 + \frac{1}{x^3} - k\right)^2 + \left(x + \frac{1}{x} - p\right)^2 = 0$

Put x = 1, then

(2 - k)² + (2 - p)² = 0

k = 2 and p = 2

Then the value of \(\frac{k}{p}\) is = \(\frac{2}{2}\) = 1

Now satisfy from the options,

If we choose $P^2-3$

$P^2-3$ = $2^2-3$ = 1 (Satisfied)

So the answer is $P^2-3$