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:

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

It will be zero, if the individual terms will be zero.

So, $$(x^3 + \frac{1}{x^3} - k)^2=0$$ and $$(x + \frac{1}{x} - p)^2$$

So, $$k=x^3 + \frac{1}{x^3}$$ and $$(x + \frac{1}{x} - p)^2=0$$

$$k=x^3 + \frac{1}{x^3}$$ and $$(x + \frac{1}{x} )=p$$

Now, $$(x + \frac{1}{x} )=p$$ taking cube of both side,

$$\Rightarrow (x + \frac{1}{x} )^3=p^3$$

$$\Rightarrow x^3+\dfrac{1}{x^3}+3(x+\dfrac{1}{x})=p^3$$

Now substituting the values in the above,

$$\Rightarrow k+3p=p^3$$

$$\Rightarrow k=p^3-3p$$

$$\Rightarrow \dfrac{k}{p}=p^2-3$$