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Use Pascal's Triangle to expand the binomial [tex]$(3x-4)^3$[/tex].

A. [tex]27x^3 - 108x^2 + 144x - 64[/tex]
B. [tex]27x^3 + 108x^2 + 144x - 64[/tex]
C. [tex]27x^3 - 108x^2 - 144x + 64[/tex]
D. [tex]27x^3 + 108x^2 + 144x + 64[/tex]

Answer :

To expand [tex]\((3x - 4)^3\)[/tex] using Pascal's Triangle, let's go through the steps of the binomial expansion.

### Step 1: Pascal's Triangle

First, identify the row of Pascal's Triangle that corresponds to [tex]\(n = 3\)[/tex]. For [tex]\(n = 3\)[/tex], the coefficients are:

[tex]\[ 1, 3, 3, 1 \][/tex]

### Step 2: Binomial Theorem Formula

The binomial theorem states:

[tex]\[
(a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k
\][/tex]

For our case [tex]\( (3x - 4)^3 \)[/tex], where [tex]\(a = 3x\)[/tex] and [tex]\(b = -4\)[/tex], substitute [tex]\(n = 3\)[/tex]:

[tex]\[
(3x - 4)^3 = \binom{3}{0}(3x)^3(-4)^0 + \binom{3}{1}(3x)^2(-4)^1 + \binom{3}{2}(3x)^1(-4)^2 + \binom{3}{3}(3x)^0(-4)^3
\][/tex]

### Step 3: Calculate Each Term

Substitute the values using the coefficients from Pascal's Triangle:

1. First term:

[tex]\(\binom{3}{0} = 1\)[/tex]

[tex]\((3x)^3 = 27x^3\)[/tex]

[tex]\((-4)^0 = 1\)[/tex]

So, the term is: [tex]\(1 \times 27x^3 \times 1 = 27x^3\)[/tex]

2. Second term:

[tex]\(\binom{3}{1} = 3\)[/tex]

[tex]\((3x)^2 = 9x^2\)[/tex]

[tex]\((-4)^1 = -4\)[/tex]

So, the term is: [tex]\(3 \times 9x^2 \times (-4) = -108x^2\)[/tex]

3. Third term:

[tex]\(\binom{3}{2} = 3\)[/tex]

[tex]\((3x)^1 = 3x\)[/tex]

[tex]\((-4)^2 = 16\)[/tex]

So, the term is: [tex]\(3 \times 3x \times 16 = 144x\)[/tex]

4. Fourth term:

[tex]\(\binom{3}{3} = 1\)[/tex]

[tex]\((3x)^0 = 1\)[/tex]

[tex]\((-4)^3 = -64\)[/tex]

So, the term is: [tex]\(1 \times 1 \times (-64) = -64\)[/tex]

### Step 4: Combine the Terms

Putting it all together, the expanded form of [tex]\((3x - 4)^3\)[/tex] is:

[tex]\[
27x^3 - 108x^2 + 144x - 64
\][/tex]

Thus, from the given options, the correct expansion is:

[tex]\[ 27x^3 - 108x^2 + 144x - 64 \][/tex]

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