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Question

Answer

$$(\dfrac{x}{y}+4)^2(\dfrac{x}{y}+1)-\dfrac{1}{x^3}-c=(\dfrac{x}{y}+4)^2\dfrac{x+y}{y}-\dfrac{1}{x^3}-c$$

Explanation
$$ \begin{aligned}(\frac{x}{y}+4)^2(\frac{x}{y}+1)-\frac{1}{x^3}-c& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}(\frac{x}{y}+4)^2\frac{x+y}{y}-\frac{1}{x^3}-c\end{aligned} $$
①

Add $ \dfrac{x}{y} $ and $ 1 $ to get $ \dfrac{ \color{purple}{ x+y } }{ y }$.

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

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

$$ \begin{aligned} \frac{x}{y} +1 & \xlongequal{\text{Step 1}} \frac{x}{y} + \frac{1}{\color{red}{1}} = \frac{ x }{ y } + \frac{ 1 \cdot \color{blue}{ y }}{ 1 \cdot \color{blue}{ y }} = \\[1ex] & \xlongequal{\text{Step 2}} \frac{ \color{purple}{ x } }{ y } + \frac{ \color{purple}{ y } }{ y }=\frac{ \color{purple}{ x+y } }{ y } \end{aligned} $$
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