Answer
$\frac{1}{3}\cdot \log_5{(x^2+1)}-\log_5{x+1}-log_5{x-1}$
Work Step by Step
Recall:
(1) $\sqrt[m]{a}=a^{\frac{1}{m}}$
(2) $\log_a {x^n}=n\cdot \log_a {x}$.
(3) $\log_a{xy}=\log_a{x} +\log_a{y}$
(4) $\log_a{\frac{x}{y}}=\log_a{x} -\log_a{y}$
Use Rule (4) to obtain
$\log_5{\frac{\sqrt[3] {x^2+1}}{x^2-1}}=\log_5{\sqrt[3] {x^2+1}}-\log_5{(x^2-1)}$.
Use Rule (2) to obtain
$\log_5{\sqrt[3] {x^2+1}}-\log_5{(x^2-1)}=\log_5{(x^2+1)^{\frac{1}{3}}}-\log_5{(x+1)(x-1)}.$
Use Rule (3) to obtain:
$\log_5{(x^2+1)^{\frac{1}{3}}}-\log_5{(x+1)(x-1)}=\frac{1}{3}\cdot \log_5{(x^2+1)}-\log_5{x+1}-log_5{x-1}$
