If `a , b` are two real numbers with a`<`b then a real number `c` can be found between `a and b` such that the value of `(a^2+a b+b^2)/(c^2)i s_____`

Answer

To solve the problem, we need to find a real number \( c \) that lies between \( a \) and \( b \) such that the expression

\[ \frac{a^2 + ab + b^2}{c^2} \]

is equal to a certain value. We will use the Mean Value Theorem (MVT) to derive this relationship.

### Step-by-Step Solution:

1. **Understanding the Mean Value Theorem (MVT)**:
The MVT states that if a function \( f \) is continuous on the closed interval \([a, b]\) and differentiable on the open interval \((a, b)\), then there exists at least one number \( c \) in \((a, b)\) such that:
\[ f'(c) = \frac{f(b) - f(a)}{b - a} \]

2. **Choosing a Function**:
We can define a function based on the expression we have. Let's consider:
\[ f(x) = x^3 \]
This function is continuous and differentiable everywhere.

3. **Calculating \( f(a) \) and \( f(b) \)**:
We find:
\[ f(a) = a^3 \quad \text{and} \quad f(b) = b^3 \]

4. **Applying MVT**:
According to MVT, there exists a \( c \) in \((a, b)\) such that:
\[ f'(c) = \frac{f(b) - f(a)}{b - a} \]
The derivative \( f'(x) = 3x^2 \), so we have:
\[ 3c^2 = \frac{b^3 - a^3}{b - a} \]

5. **Simplifying the Right Side**:
We can factor \( b^3 - a^3 \) using the difference of cubes formula:
\[ b^3 - a^3 = (b - a)(b^2 + ab + a^2) \]
Thus:
\[ \frac{b^3 - a^3}{b - a} = b^2 + ab + a^2 \]

6. **Setting Up the Equation**:
Substituting this back into our equation gives:
\[ 3c^2 = b^2 + ab + a^2 \]

7. **Rearranging the Equation**:
We can rearrange this to find the expression we are looking for:
\[ \frac{a^2 + ab + b^2}{c^2} = 3 \]

### Conclusion:
Thus, we conclude that the value of

\[ \frac{a^2 + ab + b^2}{c^2} \]

is equal to \( 3 \).