Answer
$$(3x-3)^3-(3x-3)^2=27x^3-90x^2+99x-36$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}(3x-3)^3-(3x-3)^2& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}27x^3-81x^2+81x-27-(9x^2-18x+9) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}27x^3-81x^2+81x-27-9x^2+18x-9 \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}27x^3-90x^2+99x-36\end{aligned} $$ | |
| ① | Find $ \left(3x-3\right)^3 $ using formula $$ (A - B) = A^3 - 3A^2B + 3AB^2 - B^3 $$where $ A = 3x $ and $ B = 3 $. $$ \left(3x-3\right)^3 = \left( 3x \right)^3-3 \cdot \left( 3x \right)^2 \cdot 3 + 3 \cdot 3x \cdot 3^2-3^3 = 27x^3-81x^2+81x-27 $$Find $ \left(3x-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}{ 3x } $ and $ B = \color{red}{ 3 }$. $$ \begin{aligned}\left(3x-3\right)^2 = \color{blue}{\left( 3x \right)^2} -2 \cdot 3x \cdot 3 + \color{red}{3^2} = 9x^2-18x+9\end{aligned} $$ |
| ② | Remove the parentheses by changing the sign of each term within them. $$ - \left( 9x^2-18x+9 \right) = -9x^2+18x-9 $$ |
| ③ | Combine like terms: $$ 27x^3 \color{blue}{-81x^2} + \color{red}{81x} \color{green}{-27} \color{blue}{-9x^2} + \color{red}{18x} \color{green}{-9} = 27x^3 \color{blue}{-90x^2} + \color{red}{99x} \color{green}{-36} $$ |
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