1.4 Simplify the following expressions fully: $$ 1.4 .1 \frac{3 x-3}{x^{2}-3 x+2} $$ $$ 1.4 .2 \frac{x^{3}-1}{x^{2}+x+1} $$ $$ 1.4 .3 \frac{16^{-x} \cdot 12^{8+1}}{3^{x} .4^{-7}} $$

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$$ \textbf{1.4.1} $$ Simplify $$ \frac{3x-3}{x^2-3x+2} $$

1. Factor the numerator:

$$
3x - 3 = 3(x-1)
$$

2. Factor the denominator:

$$
x^2 - 3x + 2 = (x-1)(x-2)
$$

3. Cancel the common factor $$(x-1)$$ (assuming $$ x \neq 1 $$):

$$
\frac{3(x-1)}{(x-1)(x-2)} = \frac{3}{x-2},\quad x\neq 1 \text{ and } x\neq 2.
$$

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$$ \textbf{1.4.2} $$ Simplify $$ \frac{x^3-1}{x^2+x+1} $$

1. Recognize that the numerator is a difference of cubes:

$$
x^3-1 = (x-1)(x^2+x+1)
$$

2. Cancel the common factor $$(x^2+x+1)$$ (assuming it is nonzero):

$$
\frac{(x-1)(x^2+x+1)}{x^2+x+1} = x-1.
$$

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$$ \textbf{1.4.3} $$ Simplify $$ \frac{16^{-x} \cdot 12^{8+1}}{3^{x} \cdot 4^{-7}} $$

1. Simplify the exponent $$8+1$$:

$$
12^{8+1} = 12^9.
$$

2. Express all numbers in terms of their prime factors:
   - $$16 = 2^4$$ so $$16^{-x} = (2^4)^{-x} = 2^{-4x}$$.
   - $$12 = 2^2 \cdot 3$$ so $$12^9 = (2^2 \cdot 3)^9 = 2^{18} \cdot 3^9$$.
   - $$4 = 2^2$$ so $$4^{-7} = (2^2)^{-7} = 2^{-14}$$.

3. Substitute these into the expression:

$$
\frac{2^{-4x} \cdot 2^{18} \cdot 3^9}{3^x \cdot 2^{-14}}
$$

4. Combine the factors with the same base:
   - For base $$2$$:

$$
2^{-4x} \cdot 2^{18} \cdot 2^{14} = 2^{-4x + 18 + 14} = 2^{32-4x}.
$$

- For base $$3$$:

$$
\frac{3^9}{3^x} = 3^{9-x}.
$$

5. Thus, the expression simplifies to:

$$
2^{32-4x}\cdot 3^{9-x}.
$$

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Final Answers:

$$
\frac{3x-3}{x^2-3x+2} = \frac{3}{x-2} \quad (x \neq 1,\; x\neq 2),
$$

$$
\frac{x^3-1}{x^2+x+1} = x-1,
$$

$$
\frac{16^{-x} \cdot 12^{9}}{3^{x} \cdot 4^{-7}} = 2^{32-4x}\cdot 3^{9-x}.
$$
$$ \frac{3x-3}{x^2-3x+2} = \frac{3}{x-2} \quad (x \neq 1,\; x\neq 2), $$ $$ \frac{x^3-1}{x^2+x+1} = x-1, $$ $$ \frac{16^{-x} \cdot 12^{9}}{3^{x} \cdot 4^{-7}} = 2^{32-4x}\cdot 3^{9-x}. $$

1.4 Simplify the following expressions fully:

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