If `(1-x+x^(2))^(4)=1+P_(1)x+P_(2)x^(2)+P_(3)x^(3)+...+P_(8)x^(8)`, then prove that : `P_(2)+P_(4)+P_(6)+P_(8)=40` and `P_(1)+P_(3)+P_(5)+P_(7)=-40`.

Explanation

To solve the problem, we need to expand the expression \((1 - x + x^2)^4\) using the Binomial Theorem and then extract the coefficients \(P_k\) for \(k = 0, 1, 2, \ldots, 8\). We will then prove the two required equations involving these coefficients.

### Step 1: Expand \((1 - x + x^2)^4\)

Using the multinomial expansion, we can express \((1 - x + x^2)^4\) as:

\[ (1 - x + x^2)^4 = \sum_{i+j+k=4} \frac{4!}{i!j!k!} (1)^i (-x)^j (x^2)^k \]

where \(i\), \(j\), and \(k\) are non-negative integers representing the number of times each term is chosen.

### Step 2: Rewrite the terms

The general term in the expansion can be written as:

\[ \frac{4!}{i!j!k!} (-1)^j x^{j + 2k} \]

This means that the power of \(x\) in each term is \(j + 2k\).

### Step 3: Find the coefficients \(P_k\)

To find the coefficients \(P_k\), we need to evaluate the number of ways to choose \(i\), \(j\), and \(k\) such that \(j + 2k = k\) for \(k = 0, 1, 2, \ldots, 8\).

- For \(P_0\) (constant term): \(i = 4, j = 0, k = 0\) gives \(P_0 = 1\).
- For \(P_1\): \(j + 2k = 1\) can occur with \(j = 1, k = 0\) (4 ways) or \(j = 0, k = 1\) (0 ways), thus \(P_1 = -4\).
- For \(P_2\): \(j + 2k = 2\) can occur with \(j = 2, k = 0\) (6 ways) or \(j = 0, k = 1\) (0 ways), thus \(P_2 = 6\).
- For \(P_3\): \(j + 2k = 3\) can occur with \(j = 3, k = 0\) (4 ways) or \(j = 1, k = 1\) (12 ways), thus \(P_3 = -4 + 12 = 8\).
- For \(P_4\): \(j + 2k = 4\) can occur with \(j = 4, k = 0\) (1 way) or \(j = 2, k = 1\) (12 ways), thus \(P_4 = 1 + 12 = 13\).
- For \(P_5\): \(j + 2k = 5\) can occur with \(j = 5, k = 0\) (0 ways) or \(j = 3, k = 1\) (4 ways), thus \(P_5 = -4\).
- For \(P_6\): \(j + 2k = 6\) can occur with \(j = 6, k = 0\) (0 ways) or \(j = 4, k = 1\) (6 ways), thus \(P_6 = 6\).
- For \(P_7\): \(j + 2k = 7\) can occur with \(j = 7, k = 0\) (0 ways) or \(j = 5, k = 1\) (0 ways), thus \(P_7 = -4\).
- For \(P_8\): \(j + 2k = 8\) can occur with \(j = 8, k = 0\) (0 ways) or \(j = 6, k = 1\) (0 ways), thus \(P_8 = 1\).

### Step 4: Calculate the sums

Now we can calculate the required sums:

1. **Even indexed coefficients**:
\[ P_2 + P_4 + P_6 + P_8 = 6 + 13 + 6 + 1 = 26 \]

2. **Odd indexed coefficients**:
\[ P_1 + P_3 + P_5 + P_7 = -4 + 8 - 4 - 4 = -4 \]

### Conclusion

Thus, we have shown that:
- \(P_2 + P_4 + P_6 + P_8 = 40\) is not satisfied as per our calculations.
- \(P_1 + P_3 + P_5 + P_7 = -40\) is also not satisfied.