A function $\text{f}$ is defined by $\text{f}(x) = ax^3 + bx^2 + cx + d$. The graph of $y = \text{f}(x)$ passes through the points $(-3, 4)$ and $(1, 8)$. Given that the graph of $y = \frac{1}{\text{f}(x)}$ has a turning point at $(2, \frac{1}{4})$, find the values of $a$, $b$, $c$ and $d$.

1. Formulate equations using the coordinates of the given points on the graph.

Given $\text{f}(x) = ax^3 + bx^2 + cx + d$.
Since the graph passes through $(-3, 4)$:
$\text{f}(-3) = a(-3)^3 + b(-3)^2 + c(-3) + d = 4$
$-27a + 9b - 3c + d = 4 \quad \text{--- (1)}$

Since the graph passes through $(1, 8)$:
$\text{f}(1) = a(1)^3 + b(1)^2 + c(1) + d = 8$
$a + b + c + d = 8 \quad \text{--- (2)}$

2. Use the position of the turning point of $y = \frac{1}{\text{f}(x)}$ to formulate a third equation.

The graph of $y = \frac{1}{\text{f}(x)}$ has a turning point at $(2, \frac{1}{4})$.
This means that when $x = 2$, $y = \frac{1}{4}$.
$\frac{1}{\text{f}(2)} = \frac{1}{4} \implies \text{f}(2) = 4$

Substitute $x = 2$ into $\text{f}(x)$:
$a(2)^3 + b(2)^2 + c(2) + d = 4$
$8a + 4b + 2c + d = 4 \quad \text{--- (3)}$

3. Use the turning point property to deduce $\text{f}'(2) = 0$ and formulate the fourth equation.

At a turning point, the derivative is zero. For $y = [\text{f}(x)]^{-1}$:
$\frac{\text{d}y}{\text{d}x} = -[\text{f}(x)]^{-2} \cdot \text{f}'(x)$
For $\frac{\text{d}y}{\text{d}x} = 0$ at $x = 2$, we require $\text{f}'(2) = 0$ (since $\text{f}(2) = 4 \neq 0$).

Differentiating $\text{f}(x)$:
$\text{f}'(x) = 3ax^2 + 2bx + c$

Substitute $x = 2$ and set to zero:
$3a(2)^2 + 2b(2) + c = 0$
$12a + 4b + c = 0 \quad \text{--- (4)}$

4. Solve the system of four linear equations simultaneously.

We have the system of linear equations:
$\begin{cases} -27a + 9b - 3c + d = 4 \\ a + b + c + d = 8 \\ 8a + 4b + 2c + d = 4 \\ 12a + 4b + c = 0 \end{cases}$

Solving the system (e.g. using a graphing calculator):
$a = 1$
$b = -1$
$c = -8$
$d = 16$