Find $ \left(2r+h\right)^3 $ using formula

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

where $ A = 2r $ and $ B = h $.

$$ \left(2r+h\right)^3 = \left( 2r \right)^3+3 \cdot \left( 2r \right)^2 \cdot h + 3 \cdot 2r \cdot h^2+h^3 = 8r^3+12hr^2+6h^2r+h^3 $$

Multiply $8r$ by $ \dfrac{h^2}{8r^3+12hr^2+6h^2r+h^3} $ to get $ \dfrac{8h^2r}{h^3+6h^2r+12hr^2+8r^3} $.

Step 1: Write $ 8r $ as a fraction by putting $ \color{red}{1} $ in the denominator.

Step 2: Multiply numerators and denominators.

Step 3: Simplify numerator and denominator.

$$ \begin{aligned} 8r \cdot \frac{h^2}{8r^3+12hr^2+6h^2r+h^3} & \xlongequal{\text{Step 1}} \frac{8r}{\color{red}{1}} \cdot \frac{h^2}{8r^3+12hr^2+6h^2r+h^3} = \\[1ex] & \xlongequal{\text{Step 2}} \frac{ 8r \cdot h^2 }{ 1 \cdot \left( 8r^3+12hr^2+6h^2r+h^3 \right) } \xlongequal{\text{Step 3}} \frac{ 8h^2r }{ 8r^3+12hr^2+6h^2r+h^3 } = \\[1ex] &= \frac{8h^2r}{h^3+6h^2r+12hr^2+8r^3} \end{aligned} $$