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(a) [tex]$(2x - y)^3$[/tex]

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]
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