Answer
$$\frac{x^3-9x}{3x(x-3)^2}=\frac{x+3}{3x-9}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{x^3-9x}{3x(x-3)^2}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{x^3-9x}{3x(x^2-6x+9)} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{x^3-9x}{3x^3-18x^2+27x} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{x+3}{3x-9}\end{aligned} $$ | |
| ① | Find $ \left(x-3\right)^2 $ using formula. $$ (A - B)^2 = \color{blue}{A^2} - 2 \cdot A \cdot B + \color{red}{B^2} $$where $ A = \color{blue}{ x } $ and $ B = \color{red}{ 3 }$. $$ \begin{aligned}\left(x-3\right)^2 = \color{blue}{x^2} -2 \cdot x \cdot 3 + \color{red}{3^2} = x^2-6x+9\end{aligned} $$ |
| ② | Multiply $ \color{blue}{3x} $ by $ \left( x^2-6x+9\right) $ $$ \color{blue}{3x} \cdot \left( x^2-6x+9\right) = 3x^3-18x^2+27x $$ |
| ③ | Simplify $ \dfrac{x^3-9x}{3x^3-18x^2+27x} $ to $ \dfrac{x+3}{3x-9} $. Factor both the denominator and the numerator, then cancel the common factor. $\color{blue}{x^2-3x}$. $$ \begin{aligned} \frac{x^3-9x}{3x^3-18x^2+27x} & =\frac{ \left( x+3 \right) \cdot \color{blue}{ \left( x^2-3x \right) }}{ \left( 3x-9 \right) \cdot \color{blue}{ \left( x^2-3x \right) }} = \\[1ex] &= \frac{x+3}{3x-9} \end{aligned} $$ |
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