Solution to the Problem

Part (a): Expansion of $(1 - 2x)^3$

The binomial expansion formula is given by: $(a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k$ For $(1 - 2x)^3$, we substitute $a = 1$, $b = -2x$, and $n = 3$: $(1 - 2x)^3 = \binom{3}{0}(1)^3(-2x)^0 + \binom{3}{1}(1)^2(-2x)^1 + \binom{3}{2}(1)^1(-2x)^2 + \binom{3}{3}(1)^0(-2x)^3$

Expanding each term:

• First term: $\binom{3}{0}(1)^3(-2x)^0 = 1$
• Second term: $\binom{3}{1}(1)^2(-2x)^1 = 3(-2x) = -6x$
• Third term: $\binom{3}{2}(1)^1(-2x)^2 = 3(4x^2) = 12x^2$
• Fourth term: $\binom{3}{3}(1)^0(-2x)^3 = 1(-8x^3) = -8x^3$

Thus: $(1 - 2x)^3 = 1 - 6x + 12x^2 - 8x^3$

Part (b): Expansion of $\left(1 - \frac{2}{x}\right)^3$

Here, we apply the same formula but replace $x$ with $\frac{1}{x}$ in the expansion of $(1 - 2x)^3$.

Substitute $x \to \frac{1}{x}$: $\left(1 - \frac{2}{x}\right)^3 = 1 - 6\left(\frac{1}{x}\right) + 12\left(\frac{1}{x^2}\right) - 8\left(\frac{1}{x^3}\right)$

Simplify: $\left(1 - \frac{2}{x}\right)^3 = 1 - \frac{6}{x} + \frac{12}{x^2} - \frac{8}{x^3}$

Part (c): Finding the constant term in the expansion of $(1 - 2x)^3 \cdot \left(1 - \frac{2}{x}\right)^3$

To find the constant term, we consider the product: $\text{Constant term in } (1 - 2x)^3 \cdot \left(1 - \frac{2}{x}\right)^3$ Using the expansions from (a) and (b): $(1 - 2x)^3 = 1 - 6x + 12x^2 - 8x^3$ $\left(1 - \frac{2}{x}\right)^3 = 1 - \frac{6}{x} + \frac{12}{x^2} - \frac{8}{x^3}$

Multiply the terms and identify the constant term (terms with no $x$):

• From $1 \cdot 1 = 1$
• From $(-6x) \cdot \left(-\frac{6}{x}\right) = 36$
• From $(12x^2) \cdot \left(\frac{12}{x^2}\right) = 144$
• From $(-8x^3) \cdot \left(-\frac{8}{x^3}\right) = 64$

Summing these contributions: $\text{Constant term} = 1 + 36 + 144 + 64 = 245$

Thus, the constant term is: $\boxed{245}$

Recap and Questions:

1. Do you want a deeper explanation of the binomial expansion formula?
2. Should I show how to simplify products in more detail?
3. Would you like to see a general formula for combining expansions?
4. Do you want to know how binomial coefficients are calculated?
5. Should I explain why terms with $x$ cancel out for the constant term?

Tip:

When finding the constant term in products of expansions, always focus on terms whose exponents cancel each other.