Use the distributive property.

Find $ \left(x+4\right)^2 $ using formula.

$$ (A + B)^2 = \color{blue}{A^2} + 2 \cdot A \cdot B + \color{red}{B^2} $$

where $ A = \color{blue}{ x } $ and $ B = \color{red}{ 4 }$.

$$ \begin{aligned}\left(x+4\right)^2 = \color{blue}{x^2} +2 \cdot x \cdot 4 + \color{red}{4^2} = x^2+8x+16\end{aligned} $$

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

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

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

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

Subtract $ \dfrac{x^2+2}{x^3+12x^2+48x+64} $ from $ \dfrac{x}{x^2+8x+16} $ to get $ \dfrac{ \color{purple}{ 4x-2 } }{ x^3+12x^2+48x+64 }$.

To subtract raitonal expressions, both fractions must have the same denominator.
We can create a common denominator by multiplying the first fraction by $\color{blue}{ x+4 }$.

$$ \begin{aligned} \frac{x}{x^2+8x+16} - \frac{x^2+2}{x^3+12x^2+48x+64} & = \frac{ x \cdot \color{blue}{ \left( x+4 \right) }}{ \left( x^2+8x+16 \right) \cdot \color{blue}{ \left( x+4 \right) }} - \frac{ x^2+2 }{ x^3+12x^2+48x+64 } = \\[1ex] &=\frac{ \color{purple}{ x^2+4x } }{ x^3+4x^2+8x^2+32x+16x+64 } - \frac{ \color{purple}{ x^2+2 } }{ x^3+4x^2+8x^2+32x+16x+64 } = \\[1ex] &=\frac{ \color{purple}{ x^2+4x - \left( x^2+2 \right) } }{ x^3+12x^2+48x+64 }=\frac{ \color{purple}{ 4x-2 } }{ x^3+12x^2+48x+64 } \end{aligned} $$

Multiply $ \dfrac{4x-2}{x^3+12x^2+48x+64} $ by $ 2 $ to get $ \dfrac{ 8x-4 }{ x^3+12x^2+48x+64 } $.

Step 1: Write $ 2 $ as a fraction by putting $ \color{red}{ 1 } $ in the denominator.

Step 2: Multiply numerators and denominators.

Step 3: Simplify numerator and denominator.

$$ \begin{aligned} \frac{4x-2}{x^3+12x^2+48x+64} \cdot 2 & \xlongequal{\text{Step 1}} \frac{4x-2}{x^3+12x^2+48x+64} \cdot \frac{2}{\color{red}{1}} = \\[1ex] & \xlongequal{\text{Step 2}} \frac{ \left( 4x-2 \right) \cdot 2 }{ \left( x^3+12x^2+48x+64 \right) \cdot 1 } \xlongequal{\text{Step 3}} \frac{ 8x-4 }{ x^3+12x^2+48x+64 } \end{aligned} $$