Given that $\log_a 125 - 3\log_a \sqrt{b} + \log_a c = 3$, express $a$ in terms of $b$ and $c$. Show that $5^{n+1} - 4(5^n) + 5^{n-1}$ is divisible by 2 for all positive integers of $n$.

1. Apply the power, product, and quotient laws of logarithms to combine the terms into a single logarithm.

$\log_a 125 - \log_a (\sqrt{b})^3 + \log_a c = 3$
$\log_a 125 - \log_a b^{\frac{3}{2}} + \log_a c = 3$
$\log_a \left( \frac{125c}{b^{\frac{3}{2}}} \right) = 3$

2. Convert the logarithmic equation into exponential form.

$a^3 = \frac{125c}{b^{\frac{3}{2}}}$

3. Solve for $a$ by taking the cube root of both sides.

$a = \left( \frac{125c}{b^{\frac{3}{2}}} \right)^{\frac{1}{3}}$
$a = \frac{125^{\frac{1}{3}} c^{\frac{1}{3}}}{(b^{\frac{3}{2}})^{\frac{1}{3}}}$
$a = \frac{5c^{\frac{1}{3}}}{b^{\frac{1}{2}}}$

1. Factor out the lowest power of 5, which is $5^{n-1}$.

$5^{n+1} - 4(5^n) + 5^{n-1} = 5^{n-1}(5^2 - 4(5^1) + 1)$

2. Evaluate the numerical expression inside the bracket.

$= 5^{n-1}(25 - 20 + 1)$
$= 5^{n-1}(6)$

3. Show that the result is divisible by 2.

$= 5^{n-1}(2 \times 3)$
$= 2(3 \times 5^{n-1})$
Since $n$ is a positive integer, $n-1 \ge 0$, so $3 \times 5^{n-1}$ is an integer. Therefore, the expression has a factor of 2 and is divisible by 2.