Join the IDNLearn.com community and start finding the answers you need today. Ask your questions and receive reliable and comprehensive answers from our dedicated community of professionals.
Sagot :
To solve for [tex]\((2x - y)^3\)[/tex], we need to expand the expression using the binomial theorem, which states:
[tex]\[ (a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k \][/tex]
In this case, we have [tex]\(a = 2x\)[/tex], [tex]\(b = -y\)[/tex], and [tex]\(n = 3\)[/tex].
Let's expand [tex]\((2x - y)^3\)[/tex] step-by-step:
1. Identify our terms:
- [tex]\(a = 2x\)[/tex]
- [tex]\(b = -y\)[/tex]
- [tex]\(n = 3\)[/tex]
2. Apply the binomial theorem:
[tex]\[ (2x - y)^3 = \sum_{k=0}^{3} \binom{3}{k} (2x)^{3-k} (-y)^k \][/tex]
Now, let's calculate each term in the sum separately.
3. For [tex]\(k = 0\)[/tex]:
[tex]\[ \binom{3}{0} (2x)^{3-0} (-y)^0 = 1 \cdot (2x)^3 \cdot 1 = 8x^3 \][/tex]
4. For [tex]\(k = 1\)[/tex]:
[tex]\[ \binom{3}{1} (2x)^{3-1} (-y)^1 = 3 \cdot (2x)^2 \cdot (-y) = 3 \cdot 4x^2 \cdot (-y) = -12x^2y \][/tex]
5. For [tex]\(k = 2\)[/tex]:
[tex]\[ \binom{3}{2} (2x)^{3-2} (-y)^2 = 3 \cdot (2x)^1 \cdot y^2 = 3 \cdot 2x \cdot y^2 = 6xy^2 \][/tex]
6. For [tex]\(k = 3\)[/tex]:
[tex]\[ \binom{3}{3} (2x)^{3-3} (-y)^3 = 1 \cdot (2x)^0 \cdot (-y)^3 = -y^3 \][/tex]
7. Finally, sum all these terms together:
[tex]\[ (2x - y)^3 = 8x^3 - 12x^2y + 6xy^2 - y^3 \][/tex]
So, the fully expanded form of [tex]\((2x - y)^3\)[/tex] is:
[tex]\[ 8x^3 - 12x^2y + 6xy^2 - y^3 \][/tex]
[tex]\[ (a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k \][/tex]
In this case, we have [tex]\(a = 2x\)[/tex], [tex]\(b = -y\)[/tex], and [tex]\(n = 3\)[/tex].
Let's expand [tex]\((2x - y)^3\)[/tex] step-by-step:
1. Identify our terms:
- [tex]\(a = 2x\)[/tex]
- [tex]\(b = -y\)[/tex]
- [tex]\(n = 3\)[/tex]
2. Apply the binomial theorem:
[tex]\[ (2x - y)^3 = \sum_{k=0}^{3} \binom{3}{k} (2x)^{3-k} (-y)^k \][/tex]
Now, let's calculate each term in the sum separately.
3. For [tex]\(k = 0\)[/tex]:
[tex]\[ \binom{3}{0} (2x)^{3-0} (-y)^0 = 1 \cdot (2x)^3 \cdot 1 = 8x^3 \][/tex]
4. For [tex]\(k = 1\)[/tex]:
[tex]\[ \binom{3}{1} (2x)^{3-1} (-y)^1 = 3 \cdot (2x)^2 \cdot (-y) = 3 \cdot 4x^2 \cdot (-y) = -12x^2y \][/tex]
5. For [tex]\(k = 2\)[/tex]:
[tex]\[ \binom{3}{2} (2x)^{3-2} (-y)^2 = 3 \cdot (2x)^1 \cdot y^2 = 3 \cdot 2x \cdot y^2 = 6xy^2 \][/tex]
6. For [tex]\(k = 3\)[/tex]:
[tex]\[ \binom{3}{3} (2x)^{3-3} (-y)^3 = 1 \cdot (2x)^0 \cdot (-y)^3 = -y^3 \][/tex]
7. Finally, sum all these terms together:
[tex]\[ (2x - y)^3 = 8x^3 - 12x^2y + 6xy^2 - y^3 \][/tex]
So, the fully expanded form of [tex]\((2x - y)^3\)[/tex] is:
[tex]\[ 8x^3 - 12x^2y + 6xy^2 - y^3 \][/tex]
Thank you for contributing to our discussion. Don't forget to check back for new answers. Keep asking, answering, and sharing useful information. IDNLearn.com provides the best answers to your questions. Thank you for visiting, and come back soon for more helpful information.