Exponent ain't that tough

The reciprocals give us a clue that some convenient cancellation will occur when we exponentiate (using the binomial theorem).

$\begin{aligned} \left( x + \frac1x \right)^3 &= 5^3 \\ x^3 + 3x^2\cdot \frac1x +3x\cdot \frac{1}{x^2} + \frac{1}{x^3} &= 125 \\ x^3 + 3x + 3\frac1x + \frac{1}{x^3} &= 125 \\ x^3 + 3(x + \frac1x) + \frac{1}{x^3} &= 125 \\ x^3 + 3(5) + \frac{1}{x^3} &= 125 \\ x^3 + \frac{1}{x^3} &= 110 = 5\cdot 22 \end{aligned}$

$\begin{aligned} \left( x + \frac1x \right)^5 &= 5^5 \\ x^5 + 5x^3 +10x + \frac{10}{x} + \frac{5}{x^3}+ \frac{1}{x^5} &= 5^5 \\ x^5 + 5(x^3 + \frac{1}{x^3})+10(x + \frac{1}{x}) + \frac{1}{x^5} &= 5^5 \\ x^5 + 5(110)+10(5) + \frac{1}{x^5} &= 5^5 \\ x^5 + \frac{1}{x^5} &= 5^5 - 5^2(24) = 5^2(5^3 - 24) = 25\cdot 101 \\ \end{aligned}$

$\begin{aligned} \dfrac{x^5 + \frac1{x^5}}{x^3 + \frac1{x^3}} = \frac{25\cdot 101}{5 \cdot 22} = \frac{505}{22} \end{aligned}$

The option given is incorrect. It should be \frac{505}{22} instead of \frac{527}{23}.

You can also do it without the binomial theorem:

$\left(x+\frac{1}{x} \right)^2 = 5^2$ $x^2+2+\frac{1}{x^2} = 25$ $x^2+\frac{1}{x^2} = 23 \tag{1}$ $\left(x^2+\frac{1}{x^2} \right) \left(x + \frac{1}{x} \right) = 23 \cdot 5$ $x^3 + \left(x + \frac{1}{x} \right) + \frac{1}{x^3} = 115$ $x^3 + \frac{1}{x^3} = 110 \tag{2}$

Meanwhile, using equation $1$ : $\left(x^2+\frac{1}{x^2} \right)^2 = 23^2$ $x^4+2+\frac{1}{x^4} = 529$ $x^4+\frac{1}{x^4} = 527$ $\left(x^4+\frac{1}{x^4} \right) \left(x+\frac{1}{x} \right) = 527 \cdot 5$ $x^5+\left(x^3 + \frac{1}{x^3} \right) +\frac{1}{x^5} = 2635$ $x^4+\frac{1}{x^4} = 2635 - 110 = 2525$ from equation 2.

Therefore, the expression equals $\frac{2525}{110} = \frac{505}{22}$ .