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Answer :
To expand the binomial [tex]\( (3x-4)^3 \)[/tex] using Pascal's Triangle, follow these steps:
1. First, note that the expansion of [tex]\( (a+b)^3 \)[/tex] is given by
[tex]$$
a^3 + 3a^2b + 3ab^2 + b^3.
$$[/tex]
In this problem, let
[tex]$$
a = 3x \quad \text{and} \quad b = -4.
$$[/tex]
2. Substitute [tex]\( a \)[/tex] and [tex]\( b \)[/tex] into the expansion:
- Compute the first term:
[tex]$$
(3x)^3 = 27x^3.
$$[/tex]
- Compute the second term:
[tex]$$
3(3x)^2(-4) = 3 \cdot 9x^2 \cdot (-4) = -108x^2.
$$[/tex]
- Compute the third term:
[tex]$$
3(3x)(-4)^2 = 3 \cdot 3x \cdot 16 = 144x.
$$[/tex]
- Compute the fourth term:
[tex]$$
(-4)^3 = -64.
$$[/tex]
3. Combine all the terms to obtain the full expansion:
[tex]$$
(3x-4)^3 = 27x^3 - 108x^2 + 144x - 64.
$$[/tex]
Thus, the binomial [tex]\( (3x-4)^3 \)[/tex] expands to:
[tex]$$
\boxed{27x^3 - 108x^2 + 144x - 64.}
$$[/tex]
1. First, note that the expansion of [tex]\( (a+b)^3 \)[/tex] is given by
[tex]$$
a^3 + 3a^2b + 3ab^2 + b^3.
$$[/tex]
In this problem, let
[tex]$$
a = 3x \quad \text{and} \quad b = -4.
$$[/tex]
2. Substitute [tex]\( a \)[/tex] and [tex]\( b \)[/tex] into the expansion:
- Compute the first term:
[tex]$$
(3x)^3 = 27x^3.
$$[/tex]
- Compute the second term:
[tex]$$
3(3x)^2(-4) = 3 \cdot 9x^2 \cdot (-4) = -108x^2.
$$[/tex]
- Compute the third term:
[tex]$$
3(3x)(-4)^2 = 3 \cdot 3x \cdot 16 = 144x.
$$[/tex]
- Compute the fourth term:
[tex]$$
(-4)^3 = -64.
$$[/tex]
3. Combine all the terms to obtain the full expansion:
[tex]$$
(3x-4)^3 = 27x^3 - 108x^2 + 144x - 64.
$$[/tex]
Thus, the binomial [tex]\( (3x-4)^3 \)[/tex] expands to:
[tex]$$
\boxed{27x^3 - 108x^2 + 144x - 64.}
$$[/tex]
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