Key Concepts and Formulas Used:

1. Fundamental Theorem of Calculus (Part 1 and Leibniz Rule): If $g(x) = \int_{a(x)}^{b(x)} h(t) dt$, then $g'(x) = h(b(x)) \cdot b'(x) - h(a(x)) \cdot a'(x)$. For a simple upper limit $x$ and constant lower limit $a$, $g'(x) = h(x)$.
2. Variable Substitution: Used to transform a given functional equation involving a composite function into an explicit form of the function.
3. Differentiation of Power Functions: $\frac{d}{dx}(x^n) = nx^{n-1}$.
4. Properties of Summation: $\sum_{r=1}^{n} (a_r + b_r) = \sum_{r=1}^{n} a_r + \sum_{r=1}^{n} b_r$ and $\sum_{r=1}^{n} c \cdot a_r = c \sum_{r=1}^{n} a_r$.
5. Sum of First $n$ Natural Numbers: $\sum_{r=1}^{n} r = \frac{n(n+1)}{2}$.

---

Step-by-Step Solution:

1. Determine the explicit form of $g(x)$

We are given $g(x^3) = x^6 + x^7$. Our objective is to find the expression for $g(x)$ itself.

• Why this step? The definition of $g(x)$ is in terms of an integral with an upper limit of $x$. To apply the Fundamental Theorem of Calculus to this definition, we need to know the explicit form of $g(x)$, not $g$ evaluated at a different expression like $x^3$.
• Let $y = x^3$. This implies $x = y^{1/3}$.
• Substitute $y$ for $x^3$ and $y^{1/3}$ for $x$ into the given equation: $g(y) = (y^{1/3})^6 + (y^{1/3})^7$ $g(y) = y^{6/3} + y^{7/3}$ $g(y) = y^2 + y^{7/3}$
• Replacing the dummy variable $y$ with $x$, we get the explicit form of $g(x)$: $g(x) = x^2 + x^{7/3}$

2. Relate $f(x)$ to $g(x)$ using the Fundamental Theorem of Calculus

We are given that $g(x) = \int\limits_0^x t f(t) d t$.

• Why this step? The Fundamental Theorem of Calculus provides a direct link between a function defined by an integral and the integrand. Differentiating $g(x)$ with respect to $x$ will allow us to find an expression involving $f(x)$.
• Differentiate both sides of the equation with respect to $x$: $\frac{d}{dx} g(x) = \frac{d}{dx} \left( \int\limits_0^x t f(t) d t \right)$
• Using the Fundamental Theorem of Calculus (Leibniz Rule, with $a(x)=0$, $b(x)=x$, and $h(t)=tf(t)$), we get: $g'(x) = (x f(x)) \cdot \frac{d}{dx}(x) - (0 \cdot f(0)) \cdot \frac{d}{dx}(0)$ $g'(x) = x f(x) \cdot 1 - 0$ $g'(x) = x f(x)$

3. Find the explicit expression for $f(x)$

We now have the explicit form of $g(x)$ from Step 1 and the relationship between $g'(x)$ and $f(x)$ from Step 2.

• Why this step? By differentiating the explicit form of $g(x)$ and equating it to $x f(x)$, we can solve for $f(x)$.
• First, differentiate $g(x) = x^2 + x^{7/3}$ with respect to $x$: $g'(x) = \frac{d}{dx}(x^2) + \frac{d}{dx}(x^{7/3})$ $g'(x) = 2x + \frac{7}{3}x^{\frac{7}{3}-1}$ $g'(x) = 2x + \frac{7}{3}x^{\frac{4}{3}}$
• Now, substitute this into the equation $g'(x) = x f(x)$: $2x + \frac{7}{3}x^{\frac{4}{3}} = x f(x)$
• Since the function is defined on the positive real axis, $x > 0$, so $x \neq 0$. We can divide by $x$ to find $f(x)$: $f(x) = \frac{2x + \frac{7}{3}x^{\frac{4}{3}}}{x}$ $f(x) = \frac{2x}{x} + \frac{\frac{7}{3}x^{\frac{4}{3}}}{x}$ $f(x) = 2 + \frac{7}{3}x^{\frac{4}{3}-1}$ $f(x) = 2 + \frac{7}{3}x^{\frac{1}{3}}$

4. Evaluate $f(r^3)$

The problem requires us to calculate the sum $\sum_{r=1}^{15} f(r^3)$. We first need to find the expression for $f(r^3)$.

• Why this step? The summation is explicitly over $f(r^3)$, not $f(r)$. Substituting $r^3$ into the expression for $f(x)$ will give us the term needed for the summation.
• Substitute $x = r^3$ into the expression for $f(x)$: $f(r^3) = 2 + \frac{7}{3}(r^3)^{\frac{1}{3}}$ $f(r^3) = 2 + \frac{7}{3}r$

5. Calculate the sum $\sum_{r=1}^{15} f(r^3)$

Now we can compute the required sum using the expression for $f(r^3)$ derived in Step 4.

• Why this step? This is the final computation required to answer the question. $\sum_{r=1}^{15} f(r^3) = \sum_{r=1}^{15} \left(2 + \frac{7}{3}r\right)$
• Using the linearity property of summation: $\sum_{r=1}^{15} \left(2 + \frac{7}{3}r\right) = \sum_{r=1}^{15} 2 + \sum_{r=1}^{15} \frac{7}{3}r$ $= (2 \times 15) + \frac{7}{3} \sum_{r=1}^{15} r$
• Apply the formula for the sum of the first $n$ natural numbers, $\sum_{r=1}^{n} r = \frac{n(n+1)}{2}$, with $n=15$: $= 30 + \frac{7}{3} \left(\frac{15(15+1)}{2}\right)$ $= 30 + \frac{7}{3} \left(\frac{15 \times 16}{2}\right)$ $= 30 + \frac{7}{3} (15 \times 8)$ $= 30 + 7 \times \left(\frac{15}{3} \times 8\right)$ $= 30 + 7 \times (5 \times 8)$ $= 30 + 7 \times 40$ $= 30 + 280$ $= 310$

---

Common Mistakes & Tips:

1. Incorrectly handling $g(x^3)$: A common error is to assume $g(x) = x^6 + x^7$ directly or to mishandle the substitution when finding $g(x)$. Always use a clear substitution like $y=x^3$ to avoid confusion.
2. Misapplication of FTC: Ensure the Fundamental Theorem of Calculus is applied correctly, especially when the limits of integration are functions of $x$. For $\int_0^x tf(t)dt$, the derivative is $xf(x)$.
3. Algebraic Errors: Be meticulous with exponent rules during differentiation and simplification. Also, double-check arithmetic in the summation, particularly when dealing with fractions.

---

Summary:

The problem involves finding an unknown function $f(x)$ from a given integral relation involving $g(x)$ and a functional equation for $g(x^3)$. We first determined the explicit form of $g(x)$ by using a variable substitution. Then, we applied the Fundamental Theorem of Calculus to relate $g'(x)$ to $f(x)$. By differentiating $g(x)$, we found an expression for $g'(x)$ and subsequently solved for $f(x)$. Finally, we evaluated $f(r^3)$ and computed the required sum using standard summation formulas. The value of the sum is 310.

The final answer is \boxed{310} which corresponds to option (C).