Multiply each term of $ \left( \color{blue}{2a+1}\right) $ by each term in $ \left( 2b+1\right) $.

$$ \left( \color{blue}{2a+1}\right) \cdot \left( 2b+1\right) = 4ab+2a+2b+1 $$

Multiply each term of $ \left( \color{blue}{a+1}\right) $ by each term in $ \left( b+1\right) $.

$$ \left( \color{blue}{a+1}\right) \cdot \left( b+1\right) = ab+a+b+1 $$

Multiply each term of $ \left( \color{blue}{ab+a+b+1}\right) $ by each term in $ \left( c+1\right) $.

$$ \left( \color{blue}{ab+a+b+1}\right) \cdot \left( c+1\right) = abc+ab+ac+a+bc+b+c+1 $$

Combine like terms:

$$ abc+ab+ac+a+bc+b+c+1 = abc+ab+ac+bc+a+b+c+1 $$

Multiply $4ab+2a+2b+1$ by $ \dfrac{2c+1}{abc+ab+ac+bc+a+b+c+1} $ to get $ \dfrac{8abc+4ab+4ac+4bc+2a+2b+2c+1}{abc+ab+ac+bc+a+b+c+1} $.

Step 1: Write $ 4ab+2a+2b+1 $ as a fraction by putting $ \color{red}{1} $ in the denominator.

Step 2: Multiply numerators and denominators.

Step 3: Simplify numerator and denominator.

$$ \begin{aligned} 4ab+2a+2b+1 \cdot \frac{2c+1}{abc+ab+ac+bc+a+b+c+1} & \xlongequal{\text{Step 1}} \frac{4ab+2a+2b+1}{\color{red}{1}} \cdot \frac{2c+1}{abc+ab+ac+bc+a+b+c+1} = \\[1ex] & \xlongequal{\text{Step 2}} \frac{ \left( 4ab+2a+2b+1 \right) \cdot \left( 2c+1 \right) }{ 1 \cdot \left( abc+ab+ac+bc+a+b+c+1 \right) } = \\[1ex] & \xlongequal{\text{Step 3}} \frac{ 8abc+4ab+4ac+2a+4bc+2b+2c+1 }{ abc+ab+ac+bc+a+b+c+1 } = \frac{8abc+4ab+4ac+4bc+2a+2b+2c+1}{abc+ab+ac+bc+a+b+c+1} \end{aligned} $$