Sure, let's expand the expression [tex]\((x - 4y)^6\)[/tex]. To do this, we can use the Binomial Theorem, which states:
[tex]\[
(a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k
\][/tex]
For our specific problem, [tex]\( a = x \)[/tex], [tex]\( b = -4y \)[/tex], and [tex]\( n = 6 \)[/tex]. We will expand [tex]\((x - 4y)^6\)[/tex] using this theorem.
[tex]\[
(x - 4y)^6 = \sum_{k=0}^{6} \binom{6}{k} x^{6-k} (-4y)^k
\][/tex]
We need to find the binomial coefficients [tex]\(\binom{6}{k}\)[/tex] and calculate each term individually:
1. For [tex]\( k = 0 \)[/tex]:
[tex]\[
\binom{6}{0} x^{6-0} (-4y)^0 = 1 \cdot x^6 \cdot 1 = x^6
\][/tex]
2. For [tex]\( k = 1 \)[/tex]:
[tex]\[
\binom{6}{1} x^{6-1} (-4y)^1 = 6 \cdot x^5 \cdot (-4y) = -24x^5y
\][/tex]
3. For [tex]\( k = 2 \)[/tex]:
[tex]\[
\binom{6}{2} x^{6-2} (-4y)^2 = 15 \cdot x^4 \cdot (16y^2) = 240x^4y^2
\][/tex]
4. For [tex]\( k = 3 \)[/tex]:
[tex]\[
\binom{6}{3} x^{6-3} (-4y)^3 = 20 \cdot x^3 \cdot (-64y^3) = -1280x^3y^3
\][/tex]
5. For [tex]\( k = 4 \)[/tex]:
[tex]\[
\binom{6}{4} x^{6-4} (-4y)^4 = 15 \cdot x^2 \cdot (256y^4) = 3840x^2y^4
\][/tex]
6. For [tex]\( k = 5 \)[/tex]:
[tex]\[
\binom{6}{5} x^{6-5} (-4y)^5 = 6 \cdot x \cdot (-1024y^5) = -6144xy^5
\][/tex]
7. For [tex]\( k = 6 \)[/tex]:
[tex]\[
\binom{6}{6} x^{6-6} (-4y)^6 = 1 \cdot 1 \cdot 4096y^6 = 4096y^6
\][/tex]
Now, combining all these terms, we get the expanded form of [tex]\((x - 4y)^6\)[/tex]:
[tex]\[
(x - 4y)^6 = x^6 - 24x^5y + 240x^4y^2 - 1280x^3y^3 + 3840x^2y^4 - 6144xy^5 + 4096y^6
\][/tex]
So, the expansion of [tex]\((x - 4y)^6\)[/tex] is:
[tex]\[
x^6 - 24x^5y + 240x^4y^2 - 1280x^3y^3 + 3840x^2y^4 - 6144xy^5 + 4096y^6
\][/tex]
