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

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

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

$$ \left(x+y\right)^3 = x^3+3 \cdot x^2 \cdot y + 3 \cdot x \cdot y^2+y^3 = x^3+3x^2y+3xy^2+y^3 $$

Multiply $ \dfrac{1}{3} $ by $ x^3+3x^2y+3xy^2+y^3 $ to get $ \dfrac{ x^3+3x^2y+3xy^2+y^3 }{ 3 } $.

Step 1: Write $ x^3+3x^2y+3xy^2+y^3 $ as a fraction by putting $ \color{red}{ 1 } $ in the denominator.

Step 2: Multiply numerators and denominators.

Step 3: Simplify numerator and denominator.

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