If the equations `x^(5) + ax + 1= 0 and x^(6) + ax^(2) + 1= 0 ` have a common root , then the value of ' -3a ' is equal to ______.

Answer

To solve the problem, we need to find the value of \(-3a\) given that the equations \(x^5 + ax + 1 = 0\) and \(x^6 + ax^2 + 1 = 0\) have a common root. Let's denote the common root as \(\alpha\).

### Step-by-Step Solution:

1. **Substituting the common root into the first equation:**
Since \(\alpha\) is a root of the first equation, we substitute \(\alpha\) into the equation:
\[ \alpha^5 + a\alpha + 1 = 0 \tag{1} \]

2. **Substituting the common root into the second equation:**
Similarly, since \(\alpha\) is also a root of the second equation, we substitute \(\alpha\) into the second equation:
\[ \alpha^6 + a\alpha^2 + 1 = 0 \tag{2} \]

3. **Rearranging both equations:**
From equation (1), we can express \(a\) in terms of \(\alpha\):
\[ a\alpha = -\alpha^5 - 1 \implies a = \frac{-\alpha^5 - 1}{\alpha} \tag{3} \]
From equation (2), we can also express \(a\):
\[ a\alpha^2 = -\alpha^6 - 1 \implies a = \frac{-\alpha^6 - 1}{\alpha^2} \tag{4} \]

4. **Setting the two expressions for \(a\) equal:**
From equations (3) and (4), we set them equal to each other:
\[ \frac{-\alpha^5 - 1}{\alpha} = \frac{-\alpha^6 - 1}{\alpha^2} \]

5. **Cross-multiplying to eliminate the fractions:**
Cross-multiplying gives us:
\[ (-\alpha^5 - 1)\alpha^2 = (-\alpha^6 - 1)\alpha \]

6. **Expanding both sides:**
Expanding both sides results in:
\[ -\alpha^7 - \alpha^2 = -\alpha^7 - \alpha \]

7. **Simplifying the equation:**
By adding \(\alpha^7\) to both sides, we get:
\[ -\alpha^2 = -\alpha \]
This simplifies to:
\[ \alpha^2 = \alpha \]

8. **Factoring the equation:**
Factoring gives us:
\[ \alpha(\alpha - 1) = 0 \]
Thus, \(\alpha = 0\) or \(\alpha = 1\).

9. **Finding \(a\) for \(\alpha = 0\):**
If \(\alpha = 0\), substituting into equation (1):
\[ 0 + a(0) + 1 = 0 \implies 1 = 0 \quad \text{(not valid)} \]

10. **Finding \(a\) for \(\alpha = 1\):**
If \(\alpha = 1\), substituting into equation (1):
\[ 1 + a(1) + 1 = 0 \implies 1 + a + 1 = 0 \implies a + 2 = 0 \implies a = -2 \]

11. **Calculating \(-3a\):**
Now, substituting \(a = -2\) into \(-3a\):
\[ -3a = -3(-2) = 6 \]

### Final Answer:
The value of \(-3a\) is \(6\).