What is the largest integer not exceeding $8x^3 +6x - 1$, where $x =\frac12 \left(\sqrt[3]{2+\sqrt5} + \sqrt[3]{2-\sqrt5}\right)$ ?

(A): $1$, (B): $2$, (C): $3$, (D): $4$, (E) None of the above.
1. Given \( x = \frac{1}{2} \left( \sqrt[3]{2 + \sqrt{5}} + \sqrt[3]{2 - \sqrt{5}} \right) \), we need to evaluate \( 8x^3 + 6x - 1 \).

2. First, let's find \( 8x^3 \):
   \[
   8x^3 = 8 \left( \frac{1}{2} \left( \sqrt[3]{2 + \sqrt{5}} + \sqrt[3]{2 - \sqrt{5}} \right) \right)^3
   \]
   Simplifying inside the cube:
   \[
   8x^3 = 8 \left( \frac{1}{8} \left( \sqrt[3]{2 + \sqrt{5}} + \sqrt[3]{2 - \sqrt{5}} \right)^3 \right)
   \]
   \[
   8x^3 = \left( \sqrt[3]{2 + \sqrt{5}} + \sqrt[3]{2 - \sqrt{5}} \right)^3
   \]

3. Using the identity \((a + b)^3 = a^3 + b^3 + 3ab(a + b)\), let \( a = \sqrt[3]{2 + \sqrt{5}} \) and \( b = \sqrt[3]{2 - \sqrt{5}} \):
   \[
   a^3 = 2 + \sqrt{5}, \quad b^3 = 2 - \sqrt{5}
   \]
   \[
   ab = \sqrt[3]{(2 + \sqrt{5})(2 - \sqrt{5})} = \sqrt[3]{4 - 5} = \sqrt[3]{-1} = -1
   \]
   Therefore:
   \[
   (a + b)^3 = (2 + \sqrt{5}) + (2 - \sqrt{5}) + 3(-1)(a + b)
   \]
   Simplifying:
   \[
   (a + b)^3 = 4 + 3(-1)(a + b)
   \]
   \[
   (a + b)^3 = 4 - 3(a + b)
   \]

4. Since \( x = \frac{1}{2}(a + b) \), we have \( a + b = 2x \):
   \[
   (2x)^3 = 4 - 3(2x)
   \]
   \[
   8x^3 = 4 - 6x
   \]

5. Now, we need to find \( 8x^3 + 6x - 1 \):
   \[
   8x^3 + 6x - 1 = (4 - 6x) + 6x - 1
   \]
   \[
   8x^3 + 6x - 1 = 4 - 1
   \]
   \[
   8x^3 + 6x - 1 = 3
   \]

The final answer is \(\boxed{3}\).