$$\dfrac{(1-\dfrac{u^2}{c^2})^3-\dfrac{u^2}{c^2}}{1-\dfrac{u^2}{c^2}}=\dfrac{(1-\dfrac{u^2}{c^2})^3-\dfrac{u^2}{c^2}}{\dfrac{c^2-u^2}{c^2}}$$

Explanation

$$ \begin{aligned}\frac{(1-\frac{u^2}{c^2})^3-\frac{u^2}{c^2}}{1-\frac{u^2}{c^2}}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{(1-\frac{u^2}{c^2})^3-\frac{u^2}{c^2}}{\frac{c^2-u^2}{c^2}}\end{aligned} $$

①

Subtract $ \dfrac{u^2}{c^2} $ from $ 1 $ to get $ \dfrac{ \color{purple}{ c^2-u^2 } }{ c^2 }$.

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

Step 2: To subtract raitonal expressions, both fractions must have the same denominator.
We can create a common denominator by multiplying the first fraction by $\color{blue}{ c^2 }$.

$$ \begin{aligned} 1- \frac{u^2}{c^2} & \xlongequal{\text{Step 1}} \frac{1}{\color{red}{1}} - \frac{u^2}{c^2} = \frac{ 1 \cdot \color{blue}{ c^2 }}{ 1 \cdot \color{blue}{ c^2 }} - \frac{ u^2 }{ c^2 } = \\[1ex] & \xlongequal{\text{Step 2}} \frac{ \color{purple}{ c^2 } }{ c^2 } - \frac{ \color{purple}{ u^2 } }{ c^2 }=\frac{ \color{purple}{ c^2-u^2 } }{ c^2 } \end{aligned} $$

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