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Answer

$$x^2+6x-\frac{27}{x^2}+2x-63=\frac{x^4+8x^3-63x^2-27}{x^2}$$

Explanation

Tap the blue circles to see an explanation.

$$ \begin{aligned}x^2+6x-\frac{27}{x^2}+2x-63& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{x^4+6x^3-27}{x^2}+2x-63 \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{x^4+8x^3-27}{x^2}-63 \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{x^4+8x^3-63x^2-27}{x^2}\end{aligned} $$
①

Subtract $ \dfrac{27}{x^2} $ from $ x^2+6x $ to get $ \dfrac{ \color{purple}{ x^4+6x^3-27 } }{ x^2 }$.

Step 1: Write $ x^2+6x $ 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 $\color{blue}{ x^2 }$.

$$ \begin{aligned} x^2+6x- \frac{27}{x^2} & \xlongequal{\text{Step 1}} \frac{x^2+6x}{\color{red}{1}} - \frac{27}{x^2} = \frac{ \left( x^2+6x \right) \cdot \color{blue}{ x^2 }}{ 1 \cdot \color{blue}{ x^2 }} - \frac{ 27 }{ x^2 } = \\[1ex] & \xlongequal{\text{Step 2}} \frac{ \color{purple}{ x^4+6x^3 } }{ x^2 } - \frac{ \color{purple}{ 27 } }{ x^2 }=\frac{ \color{purple}{ x^4+6x^3-27 } }{ x^2 } \end{aligned} $$
②

Add $ \dfrac{x^4+6x^3-27}{x^2} $ and $ 2x $ to get $ \dfrac{ \color{purple}{ x^4+8x^3-27 } }{ x^2 }$.

Step 1: Write $ 2x $ 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 $\color{blue}{x^2}$.

$$ \begin{aligned} \frac{x^4+6x^3-27}{x^2} +2x & \xlongequal{\text{Step 1}} \frac{x^4+6x^3-27}{x^2} + \frac{2x}{\color{red}{1}} = \frac{ x^4+6x^3-27 }{ x^2 } + \frac{ 2x \cdot \color{blue}{ x^2 }}{ 1 \cdot \color{blue}{ x^2 }} = \\[1ex] & \xlongequal{\text{Step 2}} \frac{ \color{purple}{ x^4+6x^3-27 } }{ x^2 } + \frac{ \color{purple}{ 2x^3 } }{ x^2 }=\frac{ \color{purple}{ x^4+8x^3-27 } }{ x^2 } \end{aligned} $$
③

Subtract $63$ from $ \dfrac{x^4+8x^3-27}{x^2} $ to get $ \dfrac{ \color{purple}{ x^4+8x^3-63x^2-27 } }{ x^2 }$.

Step 1: Write $ 63 $ 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 second fraction by $\color{blue}{x^2}$.

$$ \begin{aligned} \frac{x^4+8x^3-27}{x^2} -63 & \xlongequal{\text{Step 1}} \frac{x^4+8x^3-27}{x^2} - \frac{63}{\color{red}{1}} = \frac{ x^4+8x^3-27 }{ x^2 } - \frac{ 63 \cdot \color{blue}{ x^2 }}{ 1 \cdot \color{blue}{ x^2 }} = \\[1ex] & \xlongequal{\text{Step 2}} \frac{ \color{purple}{ x^4+8x^3-27 } }{ x^2 } - \frac{ \color{purple}{ 63x^2 } }{ x^2 }=\frac{ \color{purple}{ x^4+8x^3-63x^2-27 } }{ x^2 } \end{aligned} $$
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