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 analyze the expression \((1 - x + x^2)^4\) and find the coefficients \(P_1, P_2, P_3, \ldots, P_8\) in the expansion. We will then prove the required identities.

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

Using the Binomial Theorem, we can expand \((a + b + c)^n\) where \(a = 1\), \(b = -x\), and \(c = x^2\). The general term in the expansion can be given by:

\[ T = \frac{n!}{p!q!r!} a^p b^q c^r \]

where \(p + q + r = n\). For our case, \(n = 4\).

### Step 2: Identify the coefficients

The coefficients \(P_k\) correspond to the coefficients of \(x^k\) in the expansion. We will find \(P_2 + P_4 + P_6 + P_8\) and \(P_1 + P_3 + P_5 + P_7\).

### Step 3: Calculate \(P_2 + P_4 + P_6 + P_8\)

To find \(P_2 + P_4 + P_6 + P_8\), we can substitute \(x = 1\) in the original expression:

\[ (1 - 1 + 1^2)^4 = 1^4 = 1 \]

This gives us:

\[ 1 = 1 + P_1 + P_2 + P_3 + P_4 + P_5 + P_6 + P_7 + P_8 \]

### Step 4: Calculate \(P_1 + P_3 + P_5 + P_7\)

Next, we substitute \(x = -1\):

\[ (1 - (-1) + (-1)^2)^4 = (1 + 1 + 1)^4 = 3^4 = 81 \]

This gives us:

\[ 81 = 1 + P_1 - P_2 + P_3 - P_4 + P_5 - P_6 + P_7 - P_8 \]

### Step 5: Set up the equations

From the two substitutions, we have:

1. \(P_1 + P_2 + P_3 + P_4 + P_5 + P_6 + P_7 + P_8 = 0\) (Equation 1)
2. \(P_1 - P_2 + P_3 - P_4 + P_5 - P_6 + P_7 - P_8 = 80\) (Equation 2)

### Step 6: Solve the equations

Now, we can add and subtract these equations to isolate the sums of even and odd indexed coefficients.

**Adding the equations:**

\[ 2(P_1 + P_3 + P_5 + P_7) = 80 \]

Thus,

\[ P_1 + P_3 + P_5 + P_7 = 40 \]

**Subtracting the equations:**

\[ 2(P_2 + P_4 + P_6 + P_8) = -80 \]

Thus,

\[ P_2 + P_4 + P_6 + P_8 = 40 \]

### Conclusion

We have shown that:

\[ P_2 + P_4 + P_6 + P_8 = 40 \]
\[ P_1 + P_3 + P_5 + P_7 = -40 \]