Find $ \left(x^2+1\right)^3 $ using formula

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

where $ A = x^2 $ and $ B = 1 $.

$$ \left(x^2+1\right)^3 = \left( x^2 \right)^3+3 \cdot \left( x^2 \right)^2 \cdot 1 + 3 \cdot x^2 \cdot 1^2+1^3 = x^6+3x^4+3x^2+1 $$

Find $ \left(x^2+1\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^2 } $ and $ B = \color{red}{ 1 }$.

$$ \begin{aligned}\left(x^2+1\right)^2 = \color{blue}{\left( x^2 \right)^2} +2 \cdot x^2 \cdot 1 + \color{red}{1^2} = x^4+2x^2+1\end{aligned} $$

Multiply $16$ by $ \dfrac{x^2}{x^6+3x^4+3x^2+1} $ to get $ \dfrac{ 16x^2 }{ x^6+3x^4+3x^2+1 } $.

Step 1: Write $ 16 $ as a fraction by putting $ \color{red}{1} $ in the denominator.

Step 2: Multiply numerators and denominators.

Step 3: Simplify numerator and denominator.

$$ \begin{aligned} 16 \cdot \frac{x^2}{x^6+3x^4+3x^2+1} & \xlongequal{\text{Step 1}} \frac{16}{\color{red}{1}} \cdot \frac{x^2}{x^6+3x^4+3x^2+1} = \\[1ex] & \xlongequal{\text{Step 2}} \frac{ 16 \cdot x^2 }{ 1 \cdot \left( x^6+3x^4+3x^2+1 \right) } \xlongequal{\text{Step 3}} \frac{ 16x^2 }{ x^6+3x^4+3x^2+1 } \end{aligned} $$

Find $ \left(x^2+1\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^2 } $ and $ B = \color{red}{ 1 }$.

$$ \begin{aligned}\left(x^2+1\right)^2 = \color{blue}{\left( x^2 \right)^2} +2 \cdot x^2 \cdot 1 + \color{red}{1^2} = x^4+2x^2+1\end{aligned} $$

Subtract $ \dfrac{4}{x^4+2x^2+1} $ from $ \dfrac{16x^2}{x^6+3x^4+3x^2+1} $ to get $ \dfrac{ \color{purple}{ 12x^2-4 } }{ x^6+3x^4+3x^2+1 }$.

To subtract raitonal expressions, both fractions must have the same denominator.
We can create a common denominator by multiplying the second fraction by $\color{blue}{x^2+1}$.

$$ \begin{aligned} \frac{16x^2}{x^6+3x^4+3x^2+1} - \frac{4}{x^4+2x^2+1} & = \frac{ 16x^2 }{ x^6+3x^4+3x^2+1 } - \frac{ 4 \cdot \color{blue}{ \left( x^2+1 \right) }}{ \left( x^4+2x^2+1 \right) \cdot \color{blue}{ \left( x^2+1 \right) }} = \\[1ex] &=\frac{ \color{purple}{ 16x^2 } }{ x^6+3x^4+3x^2+1 } - \frac{ \color{purple}{ 4x^2+4 } }{ x^6+3x^4+3x^2+1 } = \\[1ex] &=\frac{ \color{purple}{ 16x^2 - \left( 4x^2+4 \right) } }{ x^6+3x^4+3x^2+1 }=\frac{ \color{purple}{ 12x^2-4 } }{ x^6+3x^4+3x^2+1 } \end{aligned} $$