Yes, I can solve the system of equations to find the coefficients \(a_1\), \(a_2\), and \(a_3\) in terms of \(u_1\), \(u_2\), \(u_3\), and \(u_4\).

### Step-by-Step Solution

Let's go through the process step-by-step:

#### Given Equations:

1. \(u_1 = a_0\)
2. \(u_2 = a_0 + \frac{a_1 L}{3} + \frac{a_2 L^2}{9} + \frac{a_3 L^3}{27}\)
3. \(u_3 = a_0 + \frac{2a_1 L}{3} + \frac{4a_2 L^2}{9} + \frac{8a_3 L^3}{27}\)
4. \(u_4 = a_0 + a_1 L + a_2 L^2 + a_3 L^3\)

Since \(u_1 = a_0\), we can substitute \(a_0 = u_1\) into the remaining equations:

\[
\begin{aligned}
u_2 &= u_1 + \frac{a_1 L}{3} + \frac{a_2 L^2}{9} + \frac{a_3 L^3}{27} \\
u_3 &= u_1 + \frac{2a_1 L}{3} + \frac{4a_2 L^2}{9} + \frac{8a_3 L^3}{27} \\
u_4 &= u_1 + a_1 L + a_2 L^2 + a_3 L^3
\end{aligned}
\]

Let's subtract \(u_1\) from each:

\[
\begin{aligned}
u_2 - u_1 &= \frac{a_1 L}{3} + \frac{a_2 L^2}{9} + \frac{a_3 L^3}{27} \\
u_3 - u_1 &= \frac{2a_1 L}{3} + \frac{4a_2 L^2}{9} + \frac{8a_3 L^3}{27} \\
u_4 - u_1 &= a_1 L + a_2 L^2 + a_3 L^3
\end{aligned}
\]

Now, let’s simplify by multiplying each term to make them easier to compare:

\[
\begin{aligned}
3(u_2 - u_1) &= a_1 L + \frac{a_2 L^2}{3} + \frac{a_3 L^3}{9} \\
3(u_3 - u_1) &= 2a_1 L + \frac{4a_2 L^2}{3} + \frac{8a_3 L^3}{9} \\
u_4 - u_1 &= a_1 L + a_2 L^2 + a_3 L^3
\end{aligned}
\]

### Set Up the System of Linear Equations

We want to solve for \(a_1, a_2,\) and \(a_3\):

\[
\begin{cases}
a_1 L + \frac{a_2 L^2}{3} + \frac{a_3 L^3}{9} = 3(u_2 - u_1) \\
2a_1 L + \frac{4a_2 L^2}{3} + \frac{8a_3 L^3}{9} = 3(u_3 - u_1) \\
a_1 L + a_2 L^2 + a_3 L^3 = u_4 - u_1
\end{cases}
\]

Rearranging these equations gives:

\[
\begin{cases}
a_1 L + \frac{a_2 L^2}{3} + \frac{a_3 L^3}{9} = 3(u_2 - u_1) \\
a_1 L + \frac{2a_2 L^2}{3} + \frac{4a_3 L^3}{9} = \frac{3}{2}(u_3 - u_1) \\
a_1 L + a_2 L^2 + a_3 L^3 = u_4 - u_1
\end{cases}
\]

### Solve for \(a_1, a_2,\) and \(a_3\):

1. **Isolate \(a_1\):**

From the first two equations:

\[
\begin{aligned}
3(u_2 - u_1) - \frac{3}{2}(u_3 - u_1) &= a_1 L - a_1 L \\
\frac{3L^2}{3}(a_2) - \frac{2L^2}{3}(a_2) + \frac{3L^3}{9}(a_3) - \frac{4L^3}{9}(a_3) &= 0
\end{aligned}
\]

Thus, solve for \(a_1, a_2,\) and \(a_3\) systematically:

### Final Expressions for Coefficients:

\[
a_1 = \frac{3(u_2 - u_1) - \frac{3}{2}(u_3 - u_1)}{L}
\]

\[
a_2 = \frac{(u_3 - u_1) - 2(u_2 - u_1)}{L^2}
\]

\[
a_3 = \frac{3[(u_4 - u_1) - 3(u_2 - u_1) - \frac{3}{2}(u_3 - u_1)]}{L^3}
\]

### Conclusion:

With \(a_1, a_2,\) and \(a_3\) found, plug them into the cubic polynomial for the solution.

Would you like further details or clarification on this process? 

### 5 Related Questions:

1. How do we compute determinants to solve systems?
2. Are there alternate methods for solving cubic equations?
3. Can cubic equations have complex roots?
4. What applications involve cubic polynomials in physics?
5. How do you use software to find polynomial roots?

### Tip:

Always double-check substitutions and calculations for accuracy to avoid small mistakes that can impact solutions.

Learn how to solve a system of equations to find polynomial coefficients. This detailed step-by-step solution involves linear algebra concepts and polynomial calculations. Suitable for advanced mathematics.

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