Express the numerical coefficient as a power of its prime factors.

We find that $15625 = 5^6$.
So, the expression becomes $\sqrt[5]{5^6 x^5 y^8 c^{10}}$.

The simplified expression is $\boxed{\mathbf{5xyc^2 \sqrt[5]{5y^3}}}$.

b) Let R be a universal set and P and Q are subsets of R where, $n(P) = 50$, $n(Q) = 30$, $n(P \cap Q)' = 30$ and $n(P \cap Q) = 10$, what is the value of

(i) $n(P \cup Q)$ (ii) $n(R)$

b) (i) Find $n(P \cup Q)$

Substitute the given values:

The value of $n(P \cup Q)$ is $\boxed{\mathbf{70}}$.

b) (ii) Find $n(R)$

The number of elements in the complement of a set $A$ in a universal set $U$ is $n(A') = n(U) - n(A)$.
Here, $U = R$ and $A = P \cap Q$.

Substitute the given values:

The value of $n(R)$ is $\boxed{\mathbf{40}}$.

c) If $\alpha$ and $\beta$ are the roots of the equation $2x^2 - 3x + 4 = 0$. Find the value of (i) $\frac{1}{\alpha} + \frac{1}{\beta}$ (ii) $\frac{(\alpha - \beta)^2}{\alpha \beta}$

For $2x^2 - 3x + 4 = 0$, we have $a=2$, $b=-3$, $c=4$.

Sum of roots: $\alpha + \beta = -\frac{b}{a} = -\frac{(-3)}{2} = \frac{3}{2}$
Product of roots: $\alpha \beta = \frac{c}{a} = \frac{4}{2} = 2$

c) (i) Find $\frac{1}{\alpha} + \frac{1}{\beta}$

The value of $\frac{1}{\alpha} + \frac{1}{\beta}$ is $\boxed{\mathbf{\frac{3}{4}}}$.

c) (ii) Find $\frac{(\alpha - \beta)^2}{\alpha \beta}$

Substitute $\alpha + \beta = \frac{3}{2}$ and $\alpha \beta = 2$:

The value of $\frac{(\alpha - \beta)^2}{\alpha \beta}$ is $\boxed{\mathbf{-\frac{23}{8}}}$.