**Hope This Helps**
This is my attempt at a solution.

"Let f(x) be a cubic polynomial..."

Then \(\displaystyle f(x) = ax^{3} + bx^{2} + cx + d\).

"...with leading coefficient unity..."

Then \(\displaystyle a = 1\), so \(\displaystyle f(x) = x^{3} + bx^{2} + cx + d\).

"...such that \(\displaystyle f(0) = 1\)"

Then \(\displaystyle (0)^{3} + b(0)^{2} + c(0) + d = 1\).

Then, \(\displaystyle d = 1\).

Thus, we have \(\displaystyle f(x) = x^{3} + bx^{2} + cx + 1\).

"...all the roots of \(\displaystyle f'(x) = 0\) are also the roots of \(\displaystyle f(x) = 0\)."

Then we can assume \(\displaystyle f'(x)\) divides \(\displaystyle f(x)\).

Let \(\displaystyle f'(x) = 3x^{2} + 2bx + c\).

By using long division, we find:

\(\displaystyle \frac{x^{3} +bx^{2} + cx + 1}{3x^{2} + 2bx + c}=\frac{1}{3}x+ \frac{1}{9}b\) with remainder of \(\displaystyle \left ( \frac{2}{3}c - \frac{2}{9}b^{2} \right )x + 1 - \frac{1}{9}bc\)

However, since \(\displaystyle f'(x)\) divides \(\displaystyle f(x)\), the remainder will be \(\displaystyle 0\).

Thus, \(\displaystyle \left ( \frac{2}{3}c - \frac{2}{9}b^{2} \right )x + 1 - \frac{1}{9}bc = 0\)

Note that we can rewrite the right side of the equation, to get \(\displaystyle \left ( \frac{2}{3}c - \frac{2}{9}b^{2} \right )x + 1 - \frac{1}{9}bc = 0x + 0\)

Then, by using the method of corresponding coefficients, we have the following system of equations:

\begin{cases} \frac{2}{3}c - \frac{2}{9}b^{2} & = 0\\

1 - \frac{1}{9}bc & = 0 \end{cases}

Solve the first equation for \(\displaystyle c\).

Multiplying through by \(\displaystyle 9\) gives \(\displaystyle 6c - 2b^{2} = 0\)

Adding \(\displaystyle 2b^{2}\) to other side gives \(\displaystyle 6c = 2b^{2}\)

Finally, dividing through by \(\displaystyle 6\) yields \(\displaystyle c=\frac{1}{3}b^{2}\)

Substituting this into the second equation gives \(\displaystyle 1 - \frac{1}{9}b \left ( \frac{1}{3}b^{2} \right )=0\)

Which becomes \(\displaystyle 1 - \frac{1}{27}b^{3}=0\)

Multiplying by \(\displaystyle 27\) gives \(\displaystyle 27 - b^{3}=0\)

Adding \(\displaystyle b^{3}\) to both sides gives \(\displaystyle b^{3}=27\)

Taking the cube root yields \(\displaystyle b = 3\)

We can now substitute back into either equation to find that \(\displaystyle c = 3\).

We can now take the quotient from the long division process, which is a factor of \(\displaystyle f(x)\) and the found constants and rewrite the function as:

\(\displaystyle f(x) = \left( \frac{1}{3}x+\frac{1}{3} \right ) (3x^{2} +2bx +c)\)

\(\displaystyle f(x) = \left( \frac{1}{3}x+\frac{1}{3} \right ) (3x^{2} +2(3)x +(3))\)

\(\displaystyle f(x) = \left( \frac{1}{3}x+\frac{1}{3} \right ) (3x^{2} + 6x + 3)\)

Just to clean up the factors a bit, we can pull a \(\displaystyle 3\) out of the second factor and apply it to the first factor, so that it becomes:

\(\displaystyle f(x) = ( x + 1 ) (x^{2} + 2x + 1)\)

We can further factor the second factor:

\(\displaystyle f(x) = ( x + 1 ) (x + 1)(x + 1)\)

\(\displaystyle f(x) = ( x + 1 )^{3}\)

Now we can integrate to find \(\displaystyle \int (x+1)^{3}dx = \frac{1}{4}(x+1)^{4} + C\)

Therefore \(\displaystyle g(x) = \frac{1}{4}(x+1)^{4} + C\)

Now we may substitute to find:

\(\displaystyle g(3) - g(1) = \frac{1}{4}(3+1)^{4} - \frac{1}{4}(1+1)^{4}\)

\(\displaystyle g(3) - g(1) = 64 - 4\)

\(\displaystyle g(3) - g(1) = 60\)

Therefore the correct answer is C.

A couple of notes, we can drop the integration constant \(\displaystyle C\), since we are subtracting. The integration constant would be eliminated.

Also, the long division was not shown due to the difficulty of typing that up, but if you absolutely need that I can write it on paper and upload it.

Hope this helps!