Sure! Let's expand the expression [tex]\(\left(2 - \frac{x}{2} \right)^3\)[/tex] step-by-step.
We'll use the binomial theorem for expansion. The binomial theorem states that [tex]\((a + b)^n\)[/tex] can be expanded as:
[tex]\[
(a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k
\][/tex]
In our case, [tex]\(a = 2\)[/tex], [tex]\(b = -\frac{x}{2}\)[/tex], and [tex]\(n = 3\)[/tex].
1. Calculate each term of the expansion:
[tex]\[
(a + b)^3 = \sum_{k=0}^{3} \binom{3}{k} 2^{3-k} \left(-\frac{x}{2}\right)^k
\][/tex]
2. Expand each term individually:
- For [tex]\(k = 0\)[/tex]:
[tex]\[
\binom{3}{0} 2^{3-0} \left(-\frac{x}{2}\right)^0 = 1 \cdot 2^3 \cdot 1 = 8
\][/tex]
- For [tex]\(k = 1\)[/tex]:
[tex]\[
\binom{3}{1} 2^{3-1} \left(-\frac{x}{2}\right)^1 = 3 \cdot 2^2 \cdot \left(-\frac{x}{2}\right) = 3 \cdot 4 \cdot \left(-\frac{x}{2}\right) = 3 \cdot 4 \cdot \left(-\frac{x}{2}\right) = 3 \cdot (-2x) = -6x
\][/tex]
- For [tex]\(k = 2\)[/tex]:
[tex]\[
\binom{3}{2} 2^{3-2} \left(-\frac{x}{2}\right)^2 = 3 \cdot 2^1 \cdot \left(\frac{x^2}{4}\right) = 3 \cdot 2 \cdot \left(\frac{x^2}{4}\right) = 3 \cdot \frac{x^2}{2} = \frac{3}{2}x^2
\][/tex]
- For [tex]\(k = 3\)[/tex]:
[tex]\[
\binom{3}{3} 2^{3-3} \left(-\frac{x}{2}\right)^3 = 1 \cdot 2^0 \cdot \left(-\frac{x^3}{8}\right) = 1 \cdot 1 \cdot \left(-\frac{x^3}{8}\right) = -\frac{x^3}{8}
\][/tex]
3. Combine all terms to get the expanded form:
[tex]\[
8 - 6x + \frac{3}{2}x^2 - \frac{x^3}{8}
\][/tex]
Therefore, the expanded form of [tex]\(\left(2 - \frac{x}{2}\right)^3\)[/tex] is:
[tex]\[
8 - 6x + \frac{3}{2}x^2 - \frac{x^3}{8}
\][/tex]
