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To solve the problem, we need to expand the given function \(x^2 y + \sin y + e^x\) in powers of \((x - 1)\) and \((x - \pi)\). The strategy involves applying Taylor series expansion about two different points: \(x = 1\) and \(x = \pi\).

### Step 1: Expansion around \(x = 1\)

For each term in the function, we perform the Taylor expansion around \(x = 1\).

#### Term 1: \(x^2 y\)
The function \(x^2 y\) is linear in \(y\), so only \(x\) varies.
- The first few derivatives of \(x^2\) with respect to \(x\) at \(x = 1\) are:
  \[
  f(x) = x^2 \implies f'(x) = 2x \implies f''(x) = 2
  \]
Thus, the Taylor expansion of \(x^2 y\) around \(x = 1\) is:
  \[
  x^2 y = y \left( 1 + 2(x - 1) + (x - 1)^2 \right)
  \]

#### Term 2: \(\sin y\)
The function \(\sin y\) is constant with respect to \(x\), so its expansion around \(x = 1\) is just:
  \[
  \sin y
  \]

#### Term 3: \(e^x\)
The Taylor expansion of \(e^x\) around \(x = 1\) is:
  \[
  e^x = e \cdot \left( 1 + (x - 1) + \frac{(x - 1)^2}{2!} + \cdots \right)
  \]

Thus, the full expansion around \(x = 1\) becomes:
\[
x^2 y + \sin y + e^x = y \left( 1 + 2(x - 1) + (x - 1)^2 \right) + \sin y + e \left( 1 + (x - 1) + \frac{(x - 1)^2}{2} \right)
\]

### Step 2: Expansion around \(x = \pi\)

Next, we perform a similar expansion but around \(x = \pi\).

#### Term 1: \(x^2 y\)
The first few derivatives of \(x^2\) with respect to \(x\) at \(x = \pi\) are:
  \[
  f(x) = x^2 \implies f'(x) = 2x \implies f''(x) = 2
  \]
Thus, the Taylor expansion of \(x^2 y\) around \(x = \pi\) is:
  \[
  x^2 y = y \left( \pi^2 + 2\pi(x - \pi) + (x - \pi)^2 \right)
  \]

#### Term 2: \(\sin y\)
As before, \(\sin y\) remains constant, so its expansion around \(x = \pi\) is just:
  \[
  \sin y
  \]

#### Term 3: \(e^x\)
The Taylor expansion of \(e^x\) around \(x = \pi\) is:
  \[
  e^x = e^\pi \cdot \left( 1 + (x - \pi) + \frac{(x - \pi)^2}{2!} + \cdots \right)
  \]

Thus, the full expansion around \(x = \pi\) becomes:
\[
x^2 y + \sin y + e^x = y \left( \pi^2 + 2\pi(x - \pi) + (x - \pi)^2 \right) + \sin y + e^\pi \left( 1 + (x - \pi) + \frac{(x - \pi)^2}{2} \right)
\]

### Final Answer:
- The expansion around \(x = 1\):
  \[
  x^2 y + \sin y + e^x = y \left( 1 + 2(x - 1) + (x - 1)^2 \right) + \sin y + e \left( 1 + (x - 1) + \frac{(x - 1)^2}{2} \right)
  \]
- The expansion around \(x = \pi\):
  \[
  x^2 y + \sin y + e^x = y \left( \pi^2 + 2\pi(x - \pi) + (x - \pi)^2 \right) + \sin y + e^\pi \left( 1 + (x - \pi) + \frac{(x - \pi)^2}{2} \right)
  \]

Would you like further clarification or details?

### Follow-up Questions:
1. How do Taylor series expansions work for functions with two variables?
2. Can you explain the higher-order terms in the Taylor expansion?
3. What happens when we expand in powers of other points, like \(x = 0\)?
4. How can we apply these expansions to approximate functions in real-world problems?
5. What is the geometric interpretation of a Taylor expansion around a point?

### Tip:
Always check the radius of convergence of a Taylor series to ensure the expansion is valid in the region you're analyzing.

Expand (x^2 y + sin y + e^x) in powers of (x - 1) and (x - pi).

Expand the function x^2y + sin y + e^x around points x=1 and x=π using Taylor series. This involves applying the Taylor expansion for multivariable functions to both terms, with detailed steps for the expansions.

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