In the expansion of `(1 + x^(2) )^(8)` , find the difference between the coefficients of `x^(6)` and `x^(4)` .

Answer

To find the difference between the coefficients of \(x^6\) and \(x^4\) in the expansion of \((1 + x^2)^8\), we will use the Binomial Theorem.

### Step-by-Step Solution:

1. **Identify the General Term**:
The general term in the expansion of \((a + b)^n\) is given by:
\[ T_{r+1} = \binom{n}{r} a^{n-r} b^r \]
Here, \(a = 1\), \(b = x^2\), and \(n = 8\). Therefore, the general term is:
\[ T_{r+1} = \binom{8}{r} (1)^{8-r} (x^2)^r = \binom{8}{r} x^{2r} \]

2. **Find the Coefficient of \(x^6\)**:
To find the coefficient of \(x^6\), we need \(2r = 6\) which gives \(r = 3\). Thus, the coefficient of \(x^6\) is:
\[ \text{Coefficient of } x^6 = \binom{8}{3} \]

3. **Find the Coefficient of \(x^4\)**:
To find the coefficient of \(x^4\), we need \(2r = 4\) which gives \(r = 2\). Thus, the coefficient of \(x^4\) is:
\[ \text{Coefficient of } x^4 = \binom{8}{2} \]

4. **Calculate the Coefficients**:
Now we calculate \(\binom{8}{3}\) and \(\binom{8}{2}\):
\[ \binom{8}{3} = \frac{8!}{3!(8-3)!} = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = \frac{336}{6} = 56 \]
\[ \binom{8}{2} = \frac{8!}{2!(8-2)!} = \frac{8 \times 7}{2 \times 1} = \frac{56}{2} = 28 \]

5. **Find the Difference**:
The difference between the coefficients of \(x^6\) and \(x^4\) is:
\[ \text{Difference} = \binom{8}{3} - \binom{8}{2} = 56 - 28 = 28 \]

### Final Answer:
The difference between the coefficients of \(x^6\) and \(x^4\) is \(28\).