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The given mathematical expression is:

\[
\lim_{{x \to 0}} \frac{x^2 \cdot \sin(x^3) + \sin(x)}{-x + 1 - \cos(x)}
\]

To solve this limit, we will proceed step-by-step, considering some approximations and simplifications for small \(x\).

### Step-by-step approach:

1. **Taylor Series Expansions**:
   For small values of \(x\), we can use the Taylor series expansions of sine and cosine functions around 0.

   - \(\sin(x) \approx x - \frac{x^3}{6}\) for small \(x\).
   - \(\cos(x) \approx 1 - \frac{x^2}{2}\) for small \(x\).
   - \(\sin(x^3) \approx x^3\) for small \(x\) because higher-order terms are negligible for \(x \to 0\).

2. **Apply approximations**:
   Using these approximations in the numerator and denominator:

   \[
   \text{Numerator: } x^2 \cdot \sin(x^3) + \sin(x) \approx x^2 \cdot x^3 + \left(x - \frac{x^3}{6}\right) = x^5 + x - \frac{x^3}{6}
   \]
   So, the numerator simplifies to:
   \[
   x^5 + x - \frac{x^3}{6}
   \]

   \[
   \text{Denominator: } -x + 1 - \cos(x) \approx -x + 1 - \left(1 - \frac{x^2}{2}\right) = -x + \frac{x^2}{2}
   \]

3. **Simplifying the limit**:
   The limit now becomes:
   \[
   \lim_{{x \to 0}} \frac{x^5 + x - \frac{x^3}{6}}{-x + \frac{x^2}{2}}
   \]

   As \(x \to 0\), we can simplify the terms:
   - In the numerator: The dominant term is \(x\), because \(x^5\) and \(-\frac{x^3}{6}\) go to zero much faster than \(x\).
   - In the denominator: The dominant term is \(-x\), because \(\frac{x^2}{2}\) goes to zero faster than \(-x\).

   So, the limit simplifies to:
   \[
   \lim_{{x \to 0}} \frac{x}{-x} = -1
   \]

### Final Answer:
\[
\boxed{-1}
\]

---

Would you like more details on any specific part of this solution? 

Here are some related questions to further explore the topic:

1. How does the Taylor series for sine and cosine functions help in solving limits?
2. What happens to higher-order terms in the Taylor series as \(x \to 0\)?
3. Can L'Hopital's Rule be applied to this limit? If so, how would it work?
4. Why do we approximate \(\sin(x^3)\) as \(x^3\) for small \(x\)?
5. What are other common techniques for solving indeterminate forms in limits?

**Tip**: Always check if a limit results in an indeterminate form like \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\) before choosing a strategy for solving it!

Learn how to evaluate the limit of (x²sin(x³) + sin(x)) / (-x + 1 - cos(x)) as x approaches 0 using Taylor series expansions and indeterminate form simplifications. Suitable for high school calculus students.

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