Find all real solutions to $ x^3 \minus{} 3x^2 \minus{} 8x \plus{} 40 \minus{} 8\sqrt[4]{4x \plus{} 4} \equal{} 0$
1. We start with the given equation:
   \[
   x^3 - 3x^2 - 8x + 40 - 8\sqrt[4]{4x + 4} = 0
   \]

2. To simplify the term involving the fourth root, we use the Arithmetic Mean-Geometric Mean (AM-GM) inequality. The AM-GM inequality states that for any non-negative real numbers \(a_1, a_2, \ldots, a_n\),
   \[
   \frac{a_1 + a_2 + \cdots + a_n}{n} \geq \sqrt[n]{a_1 a_2 \cdots a_n}
   \]
   with equality if and only if \(a_1 = a_2 = \cdots = a_n\).

3. Applying AM-GM to the term \(\sqrt[4]{4x + 4}\), we have:
   \[
   \sqrt[4]{4x + 4} \leq \frac{4x + 4 + 48}{4} = x + 13
   \]
   This inequality holds because \(4x + 4\) is non-negative for all \(x \geq -1\).

4. Substituting this inequality into the original equation, we get:
   \[
   x^3 - 3x^2 - 8x + 40 \leq x + 13
   \]

5. Rearranging the terms, we obtain:
   \[
   x^3 - 3x^2 - 9x + 27 \leq 0
   \]

6. Factoring the left-hand side, we get:
   \[
   (x - 3)^2(x + 3) \leq 0
   \]

7. The inequality \((x - 3)^2(x + 3) \leq 0\) holds when \(x = 3\) or \((x - 3)^2 = 0\) and \(x + 3 \leq 0\). Since \((x - 3)^2 \geq 0\) for all \(x\), the only solution is:
   \[
   x = 3
   \]

8. To verify, substitute \(x = 3\) back into the original equation:
   \[
   3^3 - 3(3)^2 - 8(3) + 40 - 8\sqrt[4]{4(3) + 4} = 0
   \]
   Simplifying, we get:
   \[
   27 - 27 - 24 + 40 - 8\sqrt[4]{16} = 0
   \]
   \[
   16 - 8 \cdot 2 = 0
   \]
   \[
   16 - 16 = 0
   \]
   This confirms that \(x = 3\) is indeed a solution.

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