To expand \((x - 2y)^4\) using the Binomial Theorem, we can follow the formula:
\[ (a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k \]
In this case, we can let \(a = x\), \(b = -2y\), and \(n = 4\).
Therefore, we have:
\[ (x - 2y)^4 = (x + (-2y))^4 \]
Now, we can apply the Binomial Theorem:
\[ (x - 2y)^4 = \sum_{k=0}^{4} \binom{4}{k} x^{4-k} (-2y)^k \]
We will compute each term in the summation for \(k = 0, 1, 2, 3, 4\):
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For \(k = 0\): \[ \binom{4}{0} x^{4-0} (-2y)^0 = 1 \cdot x^4 \cdot 1 = x^4 \]
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For \(k = 1\): \[ \binom{4}{1} x^{4-1} (-2y)^1 = 4 \cdot x^3 \cdot (-2y) = -8x^3y \]
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For \(k = 2\): \[ \binom{4}{2} x^{4-2} (-2y)^2 = 6 \cdot x^2 \cdot 4y^2 = 24x^2y^2 \]
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For \(k = 3\): \[ \binom{4}{3} x^{4-3} (-2y)^3 = 4 \cdot x^1 \cdot (-8y^3) = -32xy^3 \]
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For \(k = 4\): \[ \binom{4}{4} x^{4-4} (-2y)^4 = 1 \cdot 1 \cdot 16y^4 = 16y^4 \]
Now, we combine all the terms:
\[ (x - 2y)^4 = x^4 - 8x^3y + 24x^2y^2 - 32xy^3 + 16y^4 \]
Thus, the expanded form of \((x - 2y)^4\) is:
\[ \boxed{x^4 - 8x^3y + 24x^2y^2 - 32xy^3 + 16y^4} \]