If $ p - \frac{1}{p}= 6$, then what is the value of $p^4 +\frac{1}{p^4}$ ?

1442

If x - \(\frac{1}{x}\)  = n

Then, x + \(\frac{1}{x}\)  = \(\sqrt {n^2 + 4}\)

If $ p - \frac{1}{p}= 6$

Then, p + \(\frac{1}{p}\)  = \(\sqrt {6^2 + 4}\) = \(\sqrt {40}\)

If $K+\frac{1}{K}=n$

then, $K^2+\frac{1}{K^2}$ = n² – 2

then, $p^2+\frac{1}{p^2}$ = (\(\sqrt {40}\))² – 2 = 38

and , $p^4+\frac{1}{p^4}$ = (38)² – 2 = 1444 - 2 = 1442