Expand (x^2 + 1/x^2)^6

Steps to Solve

1. Identify the Binomial Expansion Formula
   The binomial theorem states that for any positive integer $n$, $$(a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k$$
   where $\binom{n}{k}$ is the binomial coefficient.

2. Set Variables for the Expansion
   Let $a = x^2$ and $b = \frac{1}{x^2}$.
   We are expanding $(x^2 + \frac{1}{x^2})^6$, so here $n = 6$.

3. Apply the Binomial Theorem
   Using the binomial expansion, we write:
   $$(x^2 + \frac{1}{x^2})^6 = \sum_{k=0}^{6} \binom{6}{k} (x^2)^{6-k} \left(\frac{1}{x^2}\right)^k$$

4. Simplify Each Term
   Each term in the sum is:
   $$\binom{6}{k} x^{2(6-k)} \cdot \frac{1}{x^{2k}} = \binom{6}{k} x^{12 - 2k}$$
   The expansion becomes:
   $$\sum_{k=0}^{6} \binom{6}{k} x^{12 - 2k}$$

5. List Out All Terms
   Calculate $\binom{6}{k}$ for $k = 0$ to $6$:

• For $k = 0$: $\binom{6}{0} x^{12 - 0} = 1 \cdot x^{12}$
• For $k = 1$: $\binom{6}{1} x^{12 - 2} = 6 \cdot x^{10}$
• For $k = 2$: $\binom{6}{2} x^{12 - 4} = 15 \cdot x^{8}$
• For $k = 3$: $\binom{6}{3} x^{12 - 6} = 20 \cdot x^{6}$
• For $k = 4$: $\binom{6}{4} x^{12 - 8} = 15 \cdot x^{4}$
• For $k = 5$: $\binom{6}{5} x^{12 - 10} = 6 \cdot x^{2}$
• For $k = 6$: $\binom{6}{6} x^{12 - 12} = 1 \cdot x^{0} = 1$

6. Combine All Terms
   Now, combine all calculated terms: $$(x^2 + \frac{1}{x^2})^6 = x^{12} + 6x^{10} + 15x^8 + 20x^6 + 15x^4 + 6x^2 + 1$$