a cube is shrunk so that its side lengths of 2x are reduced by 3 unifs using the binomiak theorem which of the followinf is the correct expansion for the new volume of the cube

Question

GPT-4o mini · Answer

Let's start by clarifying the problem. You have a cube with side lengths of $2 x$ . The side lengths are reduced by 3 units, resulting in new side lengths of $2 x - 3$ .

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 $2 x$ , the volume $V$ is:

$V = (2 x)^{3} = 8 x^{3}$

Now, we need to find the volume of the new cube with side length $2 x - 3$ :

$V_{new} = (2 x - 3)^{3}$

To expand $(2 x - 3)^{3}$ , 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, $a = 2 x$ , $b = - 3$ , and $n = 3$ :

$(2 x - 3)^{3} = \sum_{k = 0}^{3} (\binom{3}{k}) (2 x)^{3 - k} (- 3)^{k}$

Calculating each term:

For $k = 0$ : $(\binom{3}{0}) (2 x)^{3} (- 3)^{0} = 1 \cdot (2 x)^{3} \cdot 1 = 8 x^{3}$
For $k = 1$ : $(\binom{3}{1}) (2 x)^{2} (- 3)^{1} = 3 \cdot 4 x^{2} \cdot (- 3) = - 36 x^{2}$
For $k = 2$ : $(\binom{3}{2}) (2 x)^{1} (- 3)^{2} = 3 \cdot 2 x \cdot 9 = 54 x$
For $k = 3$ : $(\binom{3}{3}) (2 x)^{0} (- 3)^{3} = 1 \cdot 1 \cdot (- 27) = - 27$

Now, adding all these terms together gives us:

$(2 x - 3)^{3} = 8 x^{3} - 36 x^{2} + 54 x - 27$

Thus, the correct expansion for the new volume of the cube is:

$8 x^{3} - 36 x^{2} + 54 x - 27$