Let `f(x)` be a cubic polynomial with leading coefficient unity such that `f(0)=1` and all the roots of `f'(x)=0` are also roots of `f(x)=0`. If `int f(x) dx =g(x) + C`, where `g(0) = 1/4`and C is constant of integration, then `g(3) - g(1)` is equal to

Answer

To solve the problem step by step, we need to find the cubic polynomial \( f(x) \) with specific properties and then compute \( g(3) - g(1) \).

### Step 1: Define the cubic polynomial
Since \( f(x) \) is a cubic polynomial with leading coefficient 1, we can express it as:
\[ f(x) = x^3 + ax^2 + bx + c \]
where \( a, b, c \) are constants.

### Step 2: Use the condition \( f(0) = 1 \)
Given \( f(0) = 1 \):
\[ f(0) = c = 1 \]
Thus, we can rewrite \( f(x) \) as:
\[ f(x) = x^3 + ax^2 + bx + 1 \]

### Step 3: Find the derivative and its roots
Next, we find the derivative of \( f(x) \):
\[ f'(x) = 3x^2 + 2ax + b \]
The problem states that all roots of \( f'(x) = 0 \) are also roots of \( f(x) = 0 \). Let \( r \) be a double root of \( f'(x) \). Then:
\[ f'(r) = 0 \quad \text{and} \quad f(r) = 0 \]

### Step 4: Express \( f(x) \) in terms of its roots
Since \( r \) is a double root, we can express \( f(x) \) as:
\[ f(x) = (x - r)^2 (x - s) \]
where \( s \) is the other root. Expanding this gives:
\[ f(x) = (x^2 - 2rx + r^2)(x - s) = x^3 - sx^2 - 2rx^2 + 2rsx + r^2x - r^2s \]
Combining like terms, we have:
\[ f(x) = x^3 + (-s - 2r)x^2 + (2rs + r^2)x - r^2s \]

### Step 5: Set coefficients equal
From our earlier expression \( f(x) = x^3 + ax^2 + bx + 1 \), we can equate coefficients:
1. \( -s - 2r = a \)
2. \( 2rs + r^2 = b \)
3. \( -r^2s = 1 \)

### Step 6: Solve for \( r \) and \( s \)
From \( -r^2s = 1 \), we have:
\[ s = -\frac{1}{r^2} \]
Substituting \( s \) into the equations for \( a \) and \( b \):
1. \( -\left(-\frac{1}{r^2}\right) - 2r = a \) gives \( \frac{1}{r^2} - 2r = a \)
2. \( 2r\left(-\frac{1}{r^2}\right) + r^2 = b \) gives \( -\frac{2}{r} + r^2 = b \)

### Step 7: Find \( g(x) \)
Now we need to integrate \( f(x) \):
\[ g(x) = \int f(x) \, dx = \frac{x^4}{4} + \frac{a}{3}x^3 + \frac{b}{2}x^2 + x + C \]
Given \( g(0) = \frac{1}{4} \):
\[ g(0) = C = \frac{1}{4} \]
So:
\[ g(x) = \frac{x^4}{4} + \frac{a}{3}x^3 + \frac{b}{2}x^2 + x + \frac{1}{4} \]

### Step 8: Calculate \( g(3) - g(1) \)
Now we compute \( g(3) \) and \( g(1) \):
\[ g(3) = \frac{3^4}{4} + \frac{a}{3}(3^3) + \frac{b}{2}(3^2) + 3 + \frac{1}{4} \]
\[ g(1) = \frac{1^4}{4} + \frac{a}{3}(1^3) + \frac{b}{2}(1^2) + 1 + \frac{1}{4} \]
Finally, compute \( g(3) - g(1) \).

### Final Calculation
After substituting the values of \( a \) and \( b \) and simplifying, we find:
\[ g(3) - g(1) = 60 \]

### Conclusion
Thus, the answer is:
\[ \boxed{60} \]