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Question

Answer

$$3x-\dfrac{9}{x^2}-6x+9=\dfrac{-3x^3+9x^2-9}{x^2}$$

Explanation

Tap the blue circles to see an explanation.

$$ \begin{aligned}3x-\frac{9}{x^2}-6x+9& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{3x^3-9}{x^2}-6x+9 \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{-3x^3-9}{x^2}+9 \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{-3x^3+9x^2-9}{x^2}\end{aligned} $$
①

Subtract $ \dfrac{9}{x^2} $ from $ 3x $ to get $ \dfrac{ \color{purple}{ 3x^3-9 } }{ x^2 }$.

Step 1: Write $ 3x $ 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 $\color{blue}{ x^2 }$.

$$ \begin{aligned} 3x- \frac{9}{x^2} & \xlongequal{\text{Step 1}} \frac{3x}{\color{red}{1}} - \frac{9}{x^2} = \frac{ 3x \cdot \color{blue}{ x^2 }}{ 1 \cdot \color{blue}{ x^2 }} - \frac{ 9 }{ x^2 } = \\[1ex] & \xlongequal{\text{Step 2}} \frac{ \color{purple}{ 3x^3 } }{ x^2 } - \frac{ \color{purple}{ 9 } }{ x^2 }=\frac{ \color{purple}{ 3x^3-9 } }{ x^2 } \end{aligned} $$
②

Subtract $6x$ from $ \dfrac{3x^3-9}{x^2} $ to get $ \dfrac{ \color{purple}{ -3x^3-9 } }{ x^2 }$.

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

$$ \begin{aligned} \frac{3x^3-9}{x^2} -6x & \xlongequal{\text{Step 1}} \frac{3x^3-9}{x^2} - \frac{6x}{\color{red}{1}} = \frac{ 3x^3-9 }{ x^2 } - \frac{ 6x \cdot \color{blue}{ x^2 }}{ 1 \cdot \color{blue}{ x^2 }} = \\[1ex] & \xlongequal{\text{Step 2}} \frac{ \color{purple}{ 3x^3-9 } }{ x^2 } - \frac{ \color{purple}{ 6x^3 } }{ x^2 }=\frac{ \color{purple}{ -3x^3-9 } }{ x^2 } \end{aligned} $$
③

Add $ \dfrac{-3x^3-9}{x^2} $ and $ 9 $ to get $ \dfrac{ \color{purple}{ -3x^3+9x^2-9 } }{ x^2 }$.

Step 1: Write $ 9 $ 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 $\color{blue}{x^2}$.

$$ \begin{aligned} \frac{-3x^3-9}{x^2} +9 & \xlongequal{\text{Step 1}} \frac{-3x^3-9}{x^2} + \frac{9}{\color{red}{1}} = \frac{ -3x^3-9 }{ x^2 } + \frac{ 9 \cdot \color{blue}{ x^2 }}{ 1 \cdot \color{blue}{ x^2 }} = \\[1ex] & \xlongequal{\text{Step 2}} \frac{ \color{purple}{ -3x^3-9 } }{ x^2 } + \frac{ \color{purple}{ 9x^2 } }{ x^2 }=\frac{ \color{purple}{ -3x^3+9x^2-9 } }{ x^2 } \end{aligned} $$
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