Find the 3 rd term of $$ (x+2)^{8} $$ $$ \frac{4 x^{6}}{28 x^{6}} $$ $$ \frac{448 x^{5}}{112 x^{6}} $$

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To find the 3rd term of the expression $$ (x+2)^{8} $$, we need to expand the expression using the binomial theorem.

The binomial theorem states that for any positive integer $$ n $$, the expansion of $$ (a+b)^n $$ is given by:
$$ (a+b)^n = \binom{n}{0} a^n b^0 + \binom{n}{1} a^{n-1} b^1 + \binom{n}{2} a^{n-2} b^2 + \ldots + \binom{n}{n-1} a^1 b^{n-1} + \binom{n}{n} a^0 b^n $$

In this case, $$ a = x $$ and $$ b = 2 $$, and we want to find the 3rd term of the expansion.

The 3rd term of the expansion is given by:
$$ \binom{8}{2} x^{8-2} 2^2 $$

Now, let's calculate the value of the 3rd term.
Simplify the expression by following steps:
- step0: Solution:
  $$ { }_{8}C_{2}x^{8-2}\times 2^{2}$$
- step1: Subtract the numbers:
  $$ { }_{8}C_{2}x^{6}\times 2^{2}$$
- step2: Expand the expression:
  $$\frac{8!}{2!\times \left(8-2\right)!}\times x^{6}\times 2^{2}$$
- step3: Subtract the numbers:
  $$\frac{8!}{2!\times 6!}\times x^{6}\times 2^{2}$$
- step4: Reduce the fraction:
  $$\frac{8\times 7\times 6\times 5\times 4\times 3}{6!}\times x^{6}\times 2^{2}$$
- step5: Multiply the terms:
  $$\frac{20160}{6!}\times x^{6}\times 2^{2}$$
- step6: Divide the terms:
  $$28x^{6}\times 2^{2}$$
- step7: Multiply the numbers:
  $$112x^{6}$$
The 3rd term of the expression $$ (x+2)^{8} $$ is $$ 112x^{6} $$.
The 3rd term of $$ (x+2)^{8} $$ is $$ 112x^{6} $$.

Find the 3 rd term of \( (x+2)^{8} \) \( \frac{4 x^{6}}{28 x^{6}} \) \( \frac{448 x^{5}}{112 x^{6}} \)

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