If $a^5 +5a^4 +10a^3 +3a^2 -9a-6 = 0$ where $a$ is a real number other than $-1$, what is $(a + 1)^3$?

$ 
\textbf{a)}\ 1
\qquad\textbf{b)}\ 3\sqrt 3
\qquad\textbf{c)}\ 7
\qquad\textbf{d)}\ 8
\qquad\textbf{e)}\ 27
$
1. Given the polynomial equation:
   \[
   a^5 + 5a^4 + 10a^3 + 3a^2 - 9a - 6 = 0
   \]
   we need to find the value of \((a + 1)^3\).

2. Notice that the polynomial can be rewritten in terms of \((a + 1)\). Let \(b = a + 1\). Then \(a = b - 1\).

3. Substitute \(a = b - 1\) into the polynomial:
   \[
   (b-1)^5 + 5(b-1)^4 + 10(b-1)^3 + 3(b-1)^2 - 9(b-1) - 6 = 0
   \]

4. Expand each term:
   \[
   (b-1)^5 = b^5 - 5b^4 + 10b^3 - 10b^2 + 5b - 1
   \]
   \[
   5(b-1)^4 = 5(b^4 - 4b^3 + 6b^2 - 4b + 1) = 5b^4 - 20b^3 + 30b^2 - 20b + 5
   \]
   \[
   10(b-1)^3 = 10(b^3 - 3b^2 + 3b - 1) = 10b^3 - 30b^2 + 30b - 10
   \]
   \[
   3(b-1)^2 = 3(b^2 - 2b + 1) = 3b^2 - 6b + 3
   \]
   \[
   -9(b-1) = -9b + 9
   \]
   \[
   -6 = -6
   \]

5. Combine all the expanded terms:
   \[
   b^5 - 5b^4 + 10b^3 - 10b^2 + 5b - 1 + 5b^4 - 20b^3 + 30b^2 - 20b + 5 + 10b^3 - 30b^2 + 30b - 10 + 3b^2 - 6b + 3 - 9b + 9 - 6 = 0
   \]

6. Simplify the equation by combining like terms:
   \[
   b^5 + (5b^4 - 5b^4) + (10b^3 - 20b^3 + 10b^3) + (-10b^2 + 30b^2 - 30b^2 + 3b^2) + (5b - 20b + 30b - 6b - 9b) + (-1 + 5 - 10 + 3 + 9 - 6) = 0
   \]
   \[
   b^5 + 0b^4 + 0b^3 - 7b^2 + 0b + 0 = 0
   \]
   \[
   b^5 - 7b^2 = 0
   \]

7. Factor the equation:
   \[
   b^2(b^3 - 7) = 0
   \]

8. Since \(b = a + 1\) and \(a \neq -1\), \(b \neq 0\). Therefore:
   \[
   b^3 - 7 = 0
   \]
   \[
   b^3 = 7
   \]

9. Thus, \((a + 1)^3 = 7\).

The final answer is \(\boxed{7}\).