If x > 0 and $x^4 + \frac{1}{x^4} = 2207$, what is the value of $( x^5 + \frac{1}{x^5})$

15127

We know that,

x⁵ + $\frac{1}{x^5}$ = (x² + $\frac{1}{x^2}$) × (x³ + $\frac{1}{x^3}$) – (x + $\frac{1}{x}$)

If x⁴ + \(\frac{1}{x^4}\) = a

then x² + \(\frac{1}{x^2}\) = \(\sqrt {a + 2}\) = b

and x + \(\frac{1}{x}\) = \(\sqrt {b + 2}\)

If x > 0

If $x^4 + \frac{1}{x^4} = 2207$

what is the value of $( x^5 + \frac{1}{x^5})$

If $x^4 + \frac{1}{x^4} = 2207$

then, x² + \(\frac{1}{x^2}\) = \(\sqrt {2207 + 2}\) = 47

and x + \(\frac{1}{x}\) = \(\sqrt {47 + 2}\) = 7

We know that If x + \(\frac{1}{x}\)  = n

then, $x^3 +\frac{1}{x^3}$ = n³ - 3 × n

 $x^3 +\frac{1}{x^3}$ = 7³ - 3 × 7 = 322

So,

x⁵ + $\frac{1}{x^5}$ = 47 × 322 – 7

x⁵ + $\frac{1}{x^5}$ = 15127