Question
Answer
$$\dfrac{2b^3-3b^2q+q^3}{(b-q)^2}=\dfrac{2b^3-3b^2q+q^3}{b^2-2bq+q^2}$$
Explanation
| $$ \begin{aligned}\frac{2b^3-3b^2q+q^3}{(b-q)^2}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{2b^3-3b^2q+q^3}{b^2-2bq+q^2}\end{aligned} $$ | |
| ① | Find $ \left(b-q\right)^2 $ using formula. $$ (A - B)^2 = \color{blue}{A^2} - 2 \cdot A \cdot B + \color{red}{B^2} $$where $ A = \color{blue}{ b } $ and $ B = \color{red}{ q }$. $$ \begin{aligned}\left(b-q\right)^2 = \color{blue}{b^2} -2 \cdot b \cdot q + \color{red}{q^2} = b^2-2bq+q^2\end{aligned} $$ |
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