Find the smallest positive integral value of a for which the greater root of the equation `x^2-(a^2+a+1)x+a(a^2+1)=0` lies between the roots of the equation `x^2-a^2x-2(a^2-2)=0`

Answer

To solve the problem, we need to find the smallest positive integral value of \( a \) such that the greater root of the equation

\[ x^2 - (a^2 + a + 1)x + a(a^2 + 1) = 0 \]

lies between the roots of the equation

\[ x^2 - a^2x - 2(a^2 - 2) = 0. \]

### Step 1: Find the greater root of the first equation

The roots of a quadratic equation \( ax^2 + bx + c = 0 \) can be found using the quadratic formula:

\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}. \]

For the first equation, we have:

- \( a = 1 \)
- \( b = -(a^2 + a + 1) \)
- \( c = a(a^2 + 1) \)

Thus, the greater root \( x_1 \) is given by:

\[ x_1 = \frac{a^2 + a + 1 + \sqrt{(a^2 + a + 1)^2 - 4 \cdot 1 \cdot a(a^2 + 1)}}{2}. \]

### Step 2: Simplify the expression under the square root

First, we need to simplify the expression inside the square root:

\[ (a^2 + a + 1)^2 - 4a(a^2 + 1). \]

Calculating \( (a^2 + a + 1)^2 \):

\[ = a^4 + 2a^3 + 3a^2 + 2a + 1. \]

Now, calculating \( 4a(a^2 + 1) \):

\[ = 4a^3 + 4a. \]

Now, substituting back:

\[ (a^2 + a + 1)^2 - 4a(a^2 + 1) = (a^4 + 2a^3 + 3a^2 + 2a + 1) - (4a^3 + 4a). \]

This simplifies to:

\[ = a^4 - 2a^3 + 3a^2 - 2a + 1. \]

### Step 3: Find the greater root

Thus, the greater root \( x_1 \) becomes:

\[ x_1 = \frac{a^2 + a + 1 + \sqrt{a^4 - 2a^3 + 3a^2 - 2a + 1}}{2}. \]

### Step 4: Find the roots of the second equation

For the second equation \( x^2 - a^2x - 2(a^2 - 2) = 0 \):

- \( a = 1 \)
- \( b = -a^2 \)
- \( c = -2(a^2 - 2) = -2a^2 + 4 \)

The roots are given by:

\[ x_2 = \frac{a^2 \pm \sqrt{(-a^2)^2 - 4 \cdot 1 \cdot (-2a^2 + 4)}}{2}. \]

Calculating the discriminant:

\[ = a^4 + 8a^2 - 16. \]

Thus, the roots are:

\[ x_2 = \frac{a^2 \pm \sqrt{a^4 + 8a^2 - 16}}{2}. \]

### Step 5: Determine the conditions for \( x_1 \) to lie between the roots \( x_2 \)

We need to ensure that:

\[ \text{min}(x_2) < x_1 < \text{max}(x_2). \]

This leads to inequalities that we need to solve.

### Step 6: Solve the inequalities

After simplifying the inequalities, we find that \( a^2 > 5 \) leads us to:

\[ a > \sqrt{5} \approx 2.236. \]

The smallest positive integral value of \( a \) that satisfies this condition is \( a = 3 \).

### Conclusion

Thus, the smallest positive integral value of \( a \) for which the greater root of the first equation lies between the roots of the second equation is:

\[ \boxed{3}. \]