Certainly! Let's expand the function [tex]\( f(x, y) = (3x - y)^6 \)[/tex].
To accomplish this, we will use the binomial theorem which states:
[tex]\[ (a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k \][/tex]
In our case [tex]\( a = 3x \)[/tex], [tex]\( b = -y \)[/tex], and [tex]\( n = 6 \)[/tex]. Therefore:
[tex]\[ f(x, y) = (3x - y)^6 = \sum_{k=0}^{6} \binom{6}{k} (3x)^{6-k} (-y)^k \][/tex]
Let's calculate each term one by one:
1. When [tex]\( k=0 \)[/tex]:
[tex]\[ \binom{6}{0} (3x)^6 (-y)^0 = 1 \cdot (3x)^6 \cdot 1 = 729x^6 \][/tex]
2. When [tex]\( k=1 \)[/tex]:
[tex]\[ \binom{6}{1} (3x)^5 (-y)^1 = 6 \cdot 3^5 x^5 \cdot (-y) = 6 \cdot 243 x^5 \cdot (-y) = -1458 x^5 y \][/tex]
3. When [tex]\( k=2 \)[/tex]:
[tex]\[ \binom{6}{2} (3x)^4 (-y)^2 = 15 \cdot 3^4 x^4 \cdot y^2 = 15 \cdot 81 x^4 \cdot y^2 = 1215 x^4 y^2 \][/tex]
4. When [tex]\( k=3 \)[/tex]:
[tex]\[ \binom{6}{3} (3x)^3 (-y)^3 = 20 \cdot 3^3 x^3 \cdot (-y)^3 = 20 \cdot 27 x^3 \cdot (-y)^3 = -540 x^3 y^3 \][/tex]
5. When [tex]\( k=4 \)[/tex]:
[tex]\[ \binom{6}{4} (3x)^2 (-y)^4 = 15 \cdot 3^2 x^2 \cdot y^4 = 15 \cdot 9 x^2 \cdot y^4 = 135 x^2 y^4 \][/tex]
6. When [tex]\( k=5 \)[/tex]:
[tex]\[ \binom{6}{5} (3x)^1 (-y)^5 = 6 \cdot 3 x \cdot (-y)^5 = 6 \cdot 3 x \cdot (-y)^5 = -18 xy^5 \][/tex]
7. When [tex]\( k=6 \)[/tex]:
[tex]\[ \binom{6}{6} (3x)^0 (-y)^6 = 1 \cdot 1 \cdot y^6 = 1 y^6 = y^6 \][/tex]
Putting all the terms together, we get the expanded form:
[tex]\[ (3x - y)^6 = 729 x^6 - 1458 x^5 y + 1215 x^4 y^2 - 540 x^3 y^3 + 135 x^2 y^4 - 18 x y^5 + y^6 \][/tex]
So, the coefficients are as follows:
[tex]\[ \boxed{729 x^6 - 1458 x^5 y + 1215 x^4 y^2 - 540 x^3 y^3 + 135 x^2 y^4 - 18 x y^5 + y^6} \][/tex]
