If the function $f$ satisfies the equation $f(x) + f\left ( \dfrac{1}{\sqrt[3]{1-x^3}}\right ) = x^3$ for every real $x \neq 1$, what is $f(-1)$?

$ 
\textbf{(A)}\ -1
\qquad\textbf{(B)}\ \dfrac 14
\qquad\textbf{(C)}\ \dfrac 12
\qquad\textbf{(D)}\ \dfrac 74
\qquad\textbf{(E)}\ \text{None of above}
$
1. Given the functional equation:
   \[
   f(x) + f\left( \frac{1}{\sqrt[3]{1-x^3}} \right) = x^3
   \]
   for every real \( x \neq 1 \).

2. Let \( a = x^3 \). Define a new function \( g \) such that \( g(a) = f(x) \). Then, the equation becomes:
   \[
   g(a) + g\left( \frac{1}{\sqrt[3]{1-a}} \right) = a
   \]

3. We need to find \( f(-1) \). Note that \( g(-1) = f(-1) \), so it suffices to find \( g(-1) \).

4. Substitute \( a = \frac{a-1}{a} \) into the equation:
   \[
   g\left( \frac{a-1}{a} \right) + g(a) = \frac{a-1}{a}
   \]
   This gives us a new equation:
   \[
   g\left( \frac{a-1}{a} \right) + g(a) = \frac{a-1}{a} \quad \text{(Equation 1)}
   \]

5. Subtract Equation 1 from the original equation:
   \[
   g\left( \frac{1}{1-a} \right) - g\left( \frac{a-1}{a} \right) = a - \frac{a-1}{a}
   \]
   Simplify the right-hand side:
   \[
   a - \frac{a-1}{a} = a - \frac{a-1}{a} = a - \frac{a}{a} + \frac{1}{a} = a - 1 + \frac{1}{a}
   \]
   So, we have:
   \[
   g\left( \frac{1}{1-a} \right) - g\left( \frac{a-1}{a} \right) = a - 1 + \frac{1}{a} \quad \text{(Equation 2)}
   \]

6. Let \( a = -1 \) in Equation 1:
   \[
   g(2) + g(-1) = 2
   \]

7. Let \( a = \frac{1}{2} \) in Equation 2:
   \[
   g(2) - g(-1) = \frac{3}{2}
   \]

8. We now have the system of equations:
   \[
   \begin{cases}
   g(2) + g(-1) = 2 \\
   g(2) - g(-1) = \frac{3}{2}
   \end{cases}
   \]

9. Add the two equations:
   \[
   (g(2) + g(-1)) + (g(2) - g(-1)) = 2 + \frac{3}{2}
   \]
   Simplify:
   \[
   2g(2) = \frac{7}{2}
   \]
   Solve for \( g(2) \):
   \[
   g(2) = \frac{7}{4}
   \]

10. Substitute \( g(2) = \frac{7}{4} \) back into the first equation:
    \[
    \frac{7}{4} + g(-1) = 2
    \]
    Solve for \( g(-1) \):
    \[
    g(-1) = 2 - \frac{7}{4} = \frac{8}{4} - \frac{7}{4} = \frac{1}{4}
    \]

Therefore, \( f(-1) = g(-1) = \boxed{\frac{1}{4}} \).