Question
Answer
$$\dfrac{(2+x)^3-8}{x}=\dfrac{x^3+6x^2+12x}{x}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{(2+x)^3-8}{x}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{8+12x+6x^2+x^3-8}{x} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{x^3+6x^2+12x}{x}\end{aligned} $$ | |
| ① | Find $ \left(2+x\right)^3 $ using formula $$ (A + B) = A^3 + 3A^2B + 3AB^2 + B^3 $$where $ A = 2 $ and $ B = x $. $$ \left(2+x\right)^3 = 2^3+3 \cdot 2^2 \cdot x + 3 \cdot 2 \cdot x^2+x^3 = 8+12x+6x^2+x^3 $$ |
| ② | Simplify numerator $$ \, \color{blue}{ \cancel{8}} \,+12x+6x^2+x^3 \, \color{blue}{ -\cancel{8}} \, = x^3+6x^2+12x $$ |
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