The polynomial $P(x)=x^3+ax+1$ has exactly one solution on the interval $[-2,0)$ and has exactly one solution on the interval $(0,1]$ where $a$ is a real number. Which of the followings cannot be equal to $P(2)$?

$ 
\textbf{(A)}\ \sqrt{17}
\qquad\textbf{(B)}\ \sqrt[3]{30}
\qquad\textbf{(C)}\ \sqrt{26}-1
\qquad\textbf{(D)}\ \sqrt {30}
\qquad\textbf{(E)}\ \sqrt [3]{10}
$
To solve the problem, we need to analyze the polynomial \( P(x) = x^3 + ax + 1 \) and determine the range of values for \( a \) such that the polynomial has exactly one solution in the intervals \([-2,0)\) and \((0,1]\). Then, we will use this information to find which of the given options cannot be equal to \( P(2) \).

1. **Analyzing the polynomial at \( x = 0 \):**
   \[
   P(0) = 0^3 + a \cdot 0 + 1 = 1
   \]
   Since \( P(0) = 1 > 0 \), the polynomial must cross the x-axis in the intervals \([-2,0)\) and \((0,1]\).

2. **Analyzing the polynomial at \( x = -2 \):**
   \[
   P(-2) = (-2)^3 + a(-2) + 1 = -8 - 2a + 1 = -7 - 2a
   \]
   For \( P(x) \) to have a root in \([-2,0)\), we need \( P(-2) < 0 \):
   \[
   -7 - 2a < 0 \implies -7 < 2a \implies a > -\frac{7}{2}
   \]

3. **Analyzing the polynomial at \( x = 1 \):**
   \[
   P(1) = 1^3 + a \cdot 1 + 1 = 1 + a + 1 = a + 2
   \]
   For \( P(x) \) to have a root in \((0,1]\), we need \( P(1) < 0 \):
   \[
   a + 2 < 0 \implies a < -2
   \]

4. **Combining the inequalities:**
   \[
   -\frac{7}{2} < a < -2
   \]

5. **Finding \( P(2) \):**
   \[
   P(2) = 2^3 + a \cdot 2 + 1 = 8 + 2a + 1 = 9 + 2a
   \]
   Using the range for \( a \):
   \[
   -\frac{7}{2} < a < -2
   \]
   Multiply by 2:
   \[
   -7 < 2a < -4
   \]
   Add 9 to each part of the inequality:
   \[
   2 < 9 + 2a < 5
   \]
   Therefore, \( 2 < P(2) < 5 \).

6. **Checking the given options:**
   - \(\sqrt{17} \approx 4.123\) (within the range \(2 < P(2) < 5\))
   - \(\sqrt[3]{30} \approx 3.107\) (within the range \(2 < P(2) < 5\))
   - \(\sqrt{26} - 1 \approx 4.099\) (within the range \(2 < P(2) < 5\))
   - \(\sqrt{30} \approx 5.477\) (not within the range \(2 < P(2) < 5\))
   - \(\sqrt[3]{10} \approx 2.154\) (within the range \(2 < P(2) < 5\))

Since \(\sqrt{30}\) is not within the range \(2 < P(2) < 5\), it cannot be equal to \( P(2) \).

The final answer is \(\boxed{\sqrt{30}}\).