Let's start by clarifying the problem. You have a cube with side lengths of \(2x\). The side lengths are reduced by 3 units, resulting in new side lengths of \(2x - 3\).
The volume of a cube is given by the formula \(V = s^3\), where \(s\) is the side length of the cube. Thus, for the original cube with side length \(2x\), the volume \(V\) is:
\[ V = (2x)^3 = 8x^3 \]
Now, we need to find the volume of the new cube with side length \(2x - 3\):
\[ V_{\text{new}} = (2x - 3)^3 \]
To expand \((2x - 3)^3\), we can use the binomial theorem, which states:
\[ (a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k \]
For our case, \(a = 2x\), \(b = -3\), and \(n = 3\):
\[ (2x - 3)^3 = \sum_{k=0}^{3} \binom{3}{k} (2x)^{3-k} (-3)^k \]
Calculating each term:
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For \(k = 0\): \[ \binom{3}{0} (2x)^{3} (-3)^{0} = 1 \cdot (2x)^{3} \cdot 1 = 8x^3 \]
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For \(k = 1\): \[ \binom{3}{1} (2x)^{2} (-3)^{1} = 3 \cdot 4x^2 \cdot (-3) = -36x^2 \]
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For \(k = 2\): \[ \binom{3}{2} (2x)^{1} (-3)^{2} = 3 \cdot 2x \cdot 9 = 54x \]
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For \(k = 3\): \[ \binom{3}{3} (2x)^{0} (-3)^{3} = 1 \cdot 1 \cdot (-27) = -27 \]
Now, adding all these terms together gives us:
\[ (2x - 3)^3 = 8x^3 - 36x^2 + 54x - 27 \]
Thus, the correct expansion for the new volume of the cube is:
\[ 8x^3 - 36x^2 + 54x - 27 \]