Find $ \left(x+4\right)^3 $ using formula

$$ (A + B) = A^3 + 3A^2B + 3AB^2 + B^3 $$

where $ A = x $ and $ B = 4 $.

$$ \left(x+4\right)^3 = x^3+3 \cdot x^2 \cdot 4 + 3 \cdot x \cdot 4^2+4^3 = x^3+12x^2+48x+64 $$

Find $ \left(x-2\right)^3 $ using formula

$$ (A - B) = A^3 - 3A^2B + 3AB^2 - B^3 $$

where $ A = x $ and $ B = 2 $.

$$ \left(x-2\right)^3 = x^3-3 \cdot x^2 \cdot 2 + 3 \cdot x \cdot 2^2-2^3 = x^3-6x^2+12x-8 $$

Multiply each term of $ \left( \color{blue}{x^3+12x^2+48x+64}\right) $ by each term in $ \left( x-7\right) $.

$$ \left( \color{blue}{x^3+12x^2+48x+64}\right) \cdot \left( x-7\right) = x^4-7x^3+12x^3-84x^2+48x^2-336x+64x-448 $$

Combine like terms:

$$ x^4 \color{blue}{-7x^3} + \color{blue}{12x^3} \color{red}{-84x^2} + \color{red}{48x^2} \color{green}{-336x} + \color{green}{64x} -448 = \\ = x^4+ \color{blue}{5x^3} \color{red}{-36x^2} \color{green}{-272x} -448 $$

Find $ \left(x-2\right)^3 $ using formula

$$ (A - B) = A^3 - 3A^2B + 3AB^2 - B^3 $$

where $ A = x $ and $ B = 2 $.

$$ \left(x-2\right)^3 = x^3-3 \cdot x^2 \cdot 2 + 3 \cdot x \cdot 2^2-2^3 = x^3-6x^2+12x-8 $$

Simplify $ \dfrac{x^4+5x^3-36x^2-272x-448}{x^2+2x-8} $ to $ \dfrac{x^3+x^2-40x-112}{x-2} $.

Factor both the denominator and the numerator, then cancel the common factor. $\color{blue}{x+4}$.

$$ \begin{aligned} \frac{x^4+5x^3-36x^2-272x-448}{x^2+2x-8} & =\frac{ \left( x^3+x^2-40x-112 \right) \cdot \color{blue}{ \left( x+4 \right) }}{ \left( x-2 \right) \cdot \color{blue}{ \left( x+4 \right) }} = \\[1ex] &= \frac{x^3+x^2-40x-112}{x-2} \end{aligned} $$

Find $ \left(x-2\right)^3 $ using formula

$$ (A - B) = A^3 - 3A^2B + 3AB^2 - B^3 $$

where $ A = x $ and $ B = 2 $.

$$ \left(x-2\right)^3 = x^3-3 \cdot x^2 \cdot 2 + 3 \cdot x \cdot 2^2-2^3 = x^3-6x^2+12x-8 $$

Divide $ \dfrac{x^3+x^2-40x-112}{x-2} $ by $ x^2-6x-7 $ to get $ \dfrac{x^2+8x+16}{x^2-x-2} $.

Step 1: To divide rational expressions, multiply the first fraction by the reciprocal of the second fraction.

Step 2: Factor numerators and denominators.

Step 3: Cancel common factors.

Step 4: Multiply numerators and denominators.

Step 5: Simplify numerator and denominator.

$$ \begin{aligned} \frac{ \frac{x^3+x^2-40x-112}{x-2} }{x^2-6x-7} & \xlongequal{\text{Step 1}} \frac{x^3+x^2-40x-112}{x-2} \cdot \frac{\color{blue}{1}}{\color{blue}{x^2-6x-7}} = \\[1ex] & \xlongequal{\text{Step 2}} \frac{ \left( x^2+8x+16 \right) \cdot \color{blue}{ \left( x-7 \right) } }{ x-2 } \cdot \frac{ 1 }{ \left( x+1 \right) \cdot \color{blue}{ \left( x-7 \right) } } = \\[1ex] & \xlongequal{\text{Step 3}} \frac{ x^2+8x+16 }{ x-2 } \cdot \frac{ 1 }{ x+1 } \xlongequal{\text{Step 4}} \frac{ \left( x^2+8x+16 \right) \cdot 1 }{ \left( x-2 \right) \cdot \left( x+1 \right) } = \\[1ex] & \xlongequal{\text{Step 5}} \frac{ x^2+8x+16 }{ x^2+x-2x-2 } = \frac{x^2+8x+16}{x^2-x-2} \end{aligned} $$

Find $ \left(x-2\right)^3 $ using formula

$$ (A - B) = A^3 - 3A^2B + 3AB^2 - B^3 $$

where $ A = x $ and $ B = 2 $.

$$ \left(x-2\right)^3 = x^3-3 \cdot x^2 \cdot 2 + 3 \cdot x \cdot 2^2-2^3 = x^3-6x^2+12x-8 $$

Multiply each term of $ \left( \color{blue}{x-4}\right) $ by each term in $ \left( x^3-6x^2+12x-8\right) $.

$$ \left( \color{blue}{x-4}\right) \cdot \left( x^3-6x^2+12x-8\right) = x^4-6x^3+12x^2-8x-4x^3+24x^2-48x+32 $$

Combine like terms:

$$ x^4 \color{blue}{-6x^3} + \color{red}{12x^2} \color{green}{-8x} \color{blue}{-4x^3} + \color{red}{24x^2} \color{green}{-48x} +32 = x^4 \color{blue}{-10x^3} + \color{red}{36x^2} \color{green}{-56x} +32 $$

Divide $ \dfrac{x^2+8x+16}{x^2-x-2} $ by $ x^4-10x^3+36x^2-56x+32 $ to get $ \dfrac{x^2+8x+16}{x^6-11x^5+44x^4-72x^3+16x^2+80x-64} $.

Step 1: To divide rational expressions, multiply the first fraction by the reciprocal of the second fraction.

Step 2: Multiply numerators and denominators.

Step 3: Simplify numerator and denominator.

$$ \begin{aligned} \frac{ \frac{x^2+8x+16}{x^2-x-2} }{x^4-10x^3+36x^2-56x+32} & \xlongequal{\text{Step 1}} \frac{x^2+8x+16}{x^2-x-2} \cdot \frac{\color{blue}{1}}{\color{blue}{x^4-10x^3+36x^2-56x+32}} = \\[1ex] & \xlongequal{\text{Step 2}} \frac{ \left( x^2+8x+16 \right) \cdot 1 }{ \left( x^2-x-2 \right) \cdot \left( x^4-10x^3+36x^2-56x+32 \right) } = \\[1ex] & \xlongequal{\text{Step 3}} \frac{ x^2+8x+16 }{ x^6-10x^5+36x^4-56x^3+32x^2-x^5+10x^4-36x^3+56x^2-32x-2x^4+20x^3-72x^2+112x-64 } = \\[1ex] &= \frac{x^2+8x+16}{x^6-11x^5+44x^4-72x^3+16x^2+80x-64} \end{aligned} $$