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:
Answer & explanation
Correct answer: option 4
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 $P^2-3$