To simplify the expression \((1 - 4r)^3\), we'll expand it using the binomial theorem. The binomial theorem states:

\[
(a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k
\]

In this case, \(a = 1\), \(b = -4r\), and \(n = 3\). So we expand \((1 - 4r)^3\) as follows:

\[
(1 - 4r)^3 = \binom{3}{0}(1)^{3}( -4r)^{0} + \binom{3}{1}(1)^{2}(-4r)^{1} + \binom{3}{2}(1)^{1}(-4r)^{2} + \binom{3}{3}(1)^{0}(-4r)^{3}
\]

Now, let's calculate each term:

1. \(\binom{3}{0} (1)^3 (-4r)^0 = 1 \times 1 \times 1 = 1\)
2. \(\binom{3}{1} (1)^2 (-4r)^1 = 3 \times 1 \times (-4r) = -12r\)
3. \(\binom{3}{2} (1)^1 (-4r)^2 = 3 \times 1 \times 16r^2 = 48r^2\)
4. \(\binom{3}{3} (1)^0 (-4r)^3 = 1 \times 1 \times -64r^3 = -64r^3\)

Adding these together:

\[
(1 - 4r)^3 = 1 - 12r + 48r^2 - 64r^3
\]

So, the expanded form of \((1 - 4r)^3\) is:

\[
1 - 12r + 48r^2 - 64r^3
\]

Would you like more details or have any questions?

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Here are 5 related questions:
1. How do you expand \((2 + 3r)^4\) using the binomial theorem?
2. What is the binomial coefficient \(\binom{4}{2}\), and how do you calculate it?
3. Can you simplify \((1 - 2x)^5\) using the binomial theorem?
4. How do you find the middle term in the expansion of \((x - y)^6\)?
5. What is the relationship between Pascal's Triangle and binomial expansions?

**Tip:** Remember that the binomial theorem is useful for expanding expressions raised to any power, and the coefficients correspond to the entries in Pascal's Triangle!

Learn how to expand the expression (1 - 4r)^3 using the binomial theorem with a detailed step-by-step solution. This problem covers key concepts in algebraic manipulation, binomial coefficients, and polynomial expansion. Suitable for advanced high school students and college algebra courses.

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